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Normal distribution

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Difficulty: 1 – 5

55629912

A random variable is modeled as \(X\sim N(70,8^2).\) a) What is the mean of \(X\)? b) What is the standard deviation of \(X\)? c) At what value is the normal density centered?

Hints

- Recall what the two parameters in \(N(\mu,\sigma^2)\) represent. - The second parameter is a variance, not a standard deviation. - A normal curve is centered at its mean.

Solution

1. In the notation \(N(\mu,\sigma^2)\), the first parameter is the mean and the second is the variance. 2. Thus, \(\mu=70\) and \(\sigma=\sqrt{8^2}=8.\) 3. A normal density is centered at its mean, so the curve is centered at \(70\).

Answer

a) \(70\). b) \(8\). c) \(70\).

55630112

Heights are modeled by a normal distribution with mean \(70\) inches and standard deviation \(5\) inches. Using the \(68\text{-}95\text{-}99.7\) rule, approximately what percentage of heights lie between \(65\) inches and \(75\) inches?

Hints

- Compare each endpoint with the mean. - Determine how many standard deviations the endpoints are from the mean. - Recall the first percentage in the \(68\text{-}95\text{-}99.7\) rule.

Solution

1. The endpoints are \(70-5=65 \qquad\text{and}\qquad 70+5=75.\) 2. Thus, the interval is one standard deviation below to one standard deviation above the mean. 3. By the Empirical Rule, approximately \(68\%\) of values lie within one standard deviation of the mean.

Answer

Approximately \(68\%\).

52512712

A normally distributed random variable \(X\) has mean \(\mu=100\) and standard deviation \(\sigma=10\). Find each probability. Round nonexact decimal probabilities to four decimal places. a) \(P(X\le100)\) b) \(P(X\le110)\) c) \(P(90\le X\le110)\) d) \(P(X>120)\) e) \(P(X=105)\)

Hints

- Use symmetry about the mean. - Convert each boundary to a standard score and use the standard normal cumulative distribution function. - Relate the interval to distances of one or two standard deviations from the mean. - A continuous random variable assigns probability \(0\) to any single exact value.

Solution

1. By symmetry, \(P(X\le100)=0.5.\) 2. For \(110\), \(z=\frac{110-100}{10}=1,\) so \(P(X\le110)=\Phi(1)\approx0.8413.\) 3. The endpoints \(90\) and \(110\) have \(z\)-scores \(-1\) and \(1\). Thus, \(P(90\le X\le110)=\Phi(1)-\Phi(-1)\approx0.6827.\) 4. For \(120\), \(z=\frac{120-100}{10}=2,\) so \(P(X>120)=1-\Phi(2)\approx0.0228.\) 5. A normal random variable is continuous, so \(P(X=105)=0.\)

Answer

a) \(0.5\) b) \(0.8413\) c) \(0.6827\) d) \(0.0228\) e) \(0\)

52513212

A normally distributed random variable \(X\) has mean \(\mu=100\) and standard deviation \(\sigma=15\). Find \(P(85\le X\le120)\). Round to four decimal places.

Hints

- Standardize each endpoint. - Use the standard normal cumulative distribution function. - Keep the endpoint inclusion unchanged when converting to z-scores.

Solution

1. Standardize the endpoints: \(z_1=\frac{85-100}{15}=-1,\qquad z_2=\frac{120-100}{15}=\frac43.\) 2. Therefore, \(P(85\le X\le120)=\Phi\left(\frac43\right)-\Phi(-1).\) 3. Using a standard normal table or technology, \(P(85\le X\le120)\approx0.7501.\)

Answer

\(P(85\le X\le120)\approx0.7501\).

52517112

For the standard normal cumulative distribution function, \(\Phi(2)\approx0.9772\). Use symmetry to find each area, reporting values to four decimal places based on the supplied approximation. a) \(P(Z\le-2)\) b) \(P(-2\le Z\le2)\) c) \(P(Z\ge2)\)

Hints

- The total area under the standard normal density is \(1\). - Use symmetry about \(0\). - The left and right tail areas beyond equal distances from the mean are equal. - Connect each area with a cumulative distribution value.

Solution

1. By symmetry, \(P(Z\le-2)=1-\Phi(2)\approx1-0.9772=0.0228.\) 2. \(P(-2\le Z\le2) =\Phi(2)-\Phi(-2) \approx0.9772-0.0228 =0.9544.\) 3. The right-tail area is \(P(Z\ge2)=1-\Phi(2)\approx0.0228.\)

Answer

a) \(0.0228\) b) \(0.9544\) c) \(0.0228\)

52517212

For a normally distributed random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), suppose \(P(\mu-2\sigma\le X\le\mu+2\sigma)\approx0.954.\) Use symmetry to estimate each probability to three decimal places. a) \(P(X\le\mu-2\sigma)\) b) \(P(X\le\mu+2\sigma)\) c) \(P(X\ge\mu+2\sigma)\)

Hints

- Find the probability outside the central interval. - Use symmetry to divide the remaining probability between the two tails. - Relate a cumulative probability to its complementary tail.

Solution

1. The probability outside the central interval is \(1-0.954=0.046.\) 2. By symmetry, each tail contains half of this probability: \(0.046/2=0.023.\) 3. Therefore, \(P(X\le\mu-2\sigma)\approx0.023.\) 4. The cumulative probability through the upper endpoint is \(1-0.023=0.977.\) 5. The right-tail probability is also approximately \(0.023\).

Answer

a) \(0.023\) b) \(0.977\) c) \(0.023\)

52517612

A random variable \(X\) is normally distributed with \(\mu=200\) and \(\sigma=50\). Find each probability, rounded to four decimal places. a) \(P(X\le275)\) b) \(P(100<X<250)\) c) \(P(X\ge150)\)

Hints

- Standardize each boundary before using the standard normal CDF. - For the interval, subtract the lower cumulative probability from the upper cumulative probability. - Use symmetry for the right-tail probability in part c).

Solution

1. Standardize using \(z=(x-200)/50\). 2. For \(275\), \(z=1.5\), so \(P(X\le275)=\Phi(1.5)\approx0.9332\). 3. For \(100\) and \(250\), the standard scores are \(-2\) and \(1\), so \(P(100<X<250)=\Phi(1)-\Phi(-2)\approx0.8186\). 4. For \(150\), \(z=-1\), so \(P(X\ge150)=1-\Phi(-1)=\Phi(1)\approx0.8413\).

Answer

a) \(0.9332\). b) \(0.8186\). c) \(0.8413\).

52518812

A random variable \(X\) is normally distributed with \(\mu=500\) and \(\sigma=20\). Find each probability. Round nonexact probabilities to four decimal places. a) \(P(480<X<520)\) b) \(P(X>550)\) c) \(P(450\le X\le490)\) d) \(P(X\le500)\)

Hints

- Standardize each boundary. - Use symmetry for intervals centered at the mean and for the probability to the left of the mean. - Use a difference of CDF values for a bounded interval.

Solution

1. Standardize using \(Z=(X-500)/20\). 2. \(P(480<X<520)=P(-1<Z<1)\approx0.6827\). 3. For \(550\), \(z=2.5\), so \(P(X>550)=1-\Phi(2.5)\approx0.0062\). 4. For \(450\) and \(490\), the standard scores are \(-2.5\) and \(-0.5\), so \(P(450\le X\le490)=\Phi(-0.5)-\Phi(-2.5)\approx0.3023\). 5. By symmetry, \(P(X\le500)=0.5\).

Answer

a) \(0.6827\). b) \(0.0062\). c) \(0.3023\). d) \(0.5\).

52519312

The fill weight \(X\), in grams, of a brand of coffee packages is normally distributed with mean \(\mu=500\,\text{g}\) and standard deviation \(\sigma=6\,\text{g}\). Interpret each event in context and find its probability. Round nonexact probabilities to four decimal places. a) \(X<491\) b) \(494\le X\le512\) c) \(X\ge500\)

Hints

- Translate each event into words before calculating. - Convert the numerical boundaries to standard scores. - Use symmetry for the event beginning at the mean.

Solution

1. Part a) means a package contains less than \(491\,\text{g}\). The standard score is \(-1.5\), so \(P(X<491)\approx0.0668\). 2. Part b) means a package contains between \(494\,\text{g}\) and \(512\,\text{g}\). The standard scores are \(-1\) and \(2\), so \(P(494\le X\le512)\approx0.8186\). 3. Part c) means a package contains at least \(500\,\text{g}\). By symmetry about the mean, \(P(X\ge500)=0.5\).

Answer

a) Less than \(491\,\text{g}\): \(0.0668\). b) Between \(494\,\text{g}\) and \(512\,\text{g}\): \(0.8186\). c) At least \(500\,\text{g}\): \(0.5\).

52519412

The height \(H\), in inches, of nine-year-old boys in a region is normally distributed with mean \(\mu=54\) inches and standard deviation \(\sigma=2\) inches. Interpret each event in context and find its probability, rounded to four decimal places. a) \(H\le50\) b) \(52<H<56\) c) \(H>58\)

Hints

- Translate the inequalities into contextual statements. - Express each boundary in standard-deviation units from the mean. - Use symmetry for equally distant tail events.

Solution

1. Part a) means a boy is at most \(50\) inches tall. Since \(z=-2\), \(P(H\le50)\approx0.0228\). 2. Part b) means a boy is between \(52\) and \(56\) inches tall. These are one standard deviation below and above the mean, so \(P(52<H<56)\approx0.6827\). 3. Part c) means a boy is taller than \(58\) inches. Since \(z=2\), \(P(H>58)\approx0.0228\).

Answer

a) At most \(50\) inches: \(0.0228\). b) Between \(52\) and \(56\) inches: \(0.6827\). c) Taller than \(58\) inches: \(0.0228\).

52520512

A random variable \(X\) is normally distributed with mean \(\mu=500\) and standard deviation \(\sigma=50\). For each interval, give another interval with exactly the same probability by using symmetry. a) \((500, 560)\) b) \((430, 480)\) c) \((525, \infty)\) d) \((380, 420)\)

Hints

- A normal density is symmetric about its mean. - Reflect a value \(x\) across \(\mu\) using \(2\mu-x\). - A right-side interval reflects to an equal-area interval on the left. - Compare the curve and intervals relative to the mean.

Solution

1. Reflect each endpoint across \(\mu=500\). The reflection of \(x\) is \(1000-x\). 2. a) \((500, 560)\) reflects to \((440, 500)\). 3. b) \((430, 480)\) reflects to \((520, 570)\). 4. c) \((525, \infty)\) reflects to \((-\infty, 475)\). 5. d) \((380, 420)\) reflects to \((580, 620)\).

Answer

a) \((440, 500)\) b) \((520, 570)\) c) \((-\infty, 475)\) d) \((580, 620)\)

52522712

The fill volume \(X\) of bottled water is modeled by a normal distribution with standard deviation \(\sigma=8\,\text{mL}\). Find the probability that a randomly selected bottle's fill volume differs from the mean \(\mu\) by at most \(12\,\text{mL}\). Round to four decimal places.

Hints

- Write the maximum-deviation condition as a symmetric interval about the mean. - Express twelve milliliters in standard-deviation units. - Use symmetry for the central standard-normal area.

Solution

1. The event is \(\mu-12\le X\le\mu+12\). 2. Standardizing gives \(-1.5\le Z\le1.5\). 3. Therefore, \(P(|X-\mu|\le12)=\Phi(1.5)-\Phi(-1.5)\approx0.8664\).

Answer

\(0.8664\).

52525612

For a normally distributed random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), find the probability that \(X\) lies within \(0.75\) standard deviation of the mean. Give your answer as a percentage rounded to two decimal places.

Hints

- Convert the stated distance from the mean directly into standard-score endpoints. - The interval is symmetric about zero after standardization. - Convert the final decimal probability to a percentage.

Solution

1. The interval is \([\mu-0.75\sigma,\mu+0.75\sigma]\). 2. Standardizing gives \(-0.75\le Z\le0.75\). 3. Thus, \(P(-0.75\le Z\le0.75)=2\Phi(0.75)-1\approx0.5467\), or \(54.67\%\).

Answer

\(54.67\%\).

52526412

A normal random variable \(Y\) has mean \(\mu=80\) and standard deviation \(\sigma=12\). Use the \(68\text{-}95\text{-}99.7\) rule to estimate each probability. a) \(P(56\le Y\le104)\) b) \(P(Y\ge116)\) c) \(P(44\le Y\le68)\)

Hints

- Express each endpoint as a number of standard deviations from the mean. - Use symmetry to split probability equally between matching tails. - For part c), subtract nested central areas and take the appropriate half.

Solution

1. The interval \([56,104]\) is \([\mu-2\sigma,\mu+2\sigma]\), so its probability is approximately \(95.4\%\). 2. The value \(116\) is \(\mu+3\sigma\). About \(0.3\%\) lies outside the central three-standard-deviation interval, so symmetry gives approximately \(0.15\%\) in the upper tail. 3. The interval \([44,68]\) is from \(\mu-3\sigma\) to \(\mu-\sigma\). Its probability is half the difference between the central \(99.7\%\) and central \(68.3\%\): \((99.7\%-68.3\%)/2=15.7\%\).

Answer

a) Approximately \(95.4\%\). b) Approximately \(0.15\%\). c) Approximately \(15.7\%\).

52526912

The weight \(X\) of flour packages filled by a machine is normally distributed with mean \(\mu=400\,\text{g}\) and standard deviation \(\sigma=8\,\text{g}\). Use the \(68\text{-}95\text{-}99.7\) rule to estimate each probability. a) A package weighs between \(392\,\text{g}\) and \(416\,\text{g}\). b) A package weighs less than \(384\,\text{g}\). c) A package's weight differs from the mean by more than \(8\,\text{g}\).

Hints

- Express every endpoint in standard-deviation units from the mean. - Combine nested central areas and use symmetry when the requested region covers only one side. - Translate “differs from the mean” into an absolute-deviation event.

Solution

1. The interval in part a) runs from \(\mu-\sigma\) to \(\mu+2\sigma\). The area is approximately \(34.15\%+47.7\%=81.85\%\). 2. The cutoff in part b) is \(\mu-2\sigma\). The two tails outside \(\mu\pm2\sigma\) total about \(4.6\%\), so the left tail is about \(2.3\%\). 3. Part c) is the complement of the central one-standard-deviation interval, so the probability is approximately \(100\%-68.3\%=31.7\%\).

Answer

a) Approximately \(81.85\%\). b) Approximately \(2.3\%\). c) Approximately \(31.7\%\).

52527012

The length \(X\) of a metal pin is normally distributed with mean \(\mu=10.00\,\text{cm}\) and standard deviation \(\sigma=0.05\,\text{cm}\). Use the \(68\text{-}95\text{-}99.7\) rule to estimate each probability. a) The length is in \([9.90\,\text{cm},10.10\,\text{cm}]\). b) The length is between \(10.10\,\text{cm}\) and \(10.15\,\text{cm}\). c) The pin is at least \(10.05\,\text{cm}\) long.

Hints

- Locate each boundary relative to \(\mu\pm\sigma\), \(\mu\pm2\sigma\), and \(\mu\pm3\sigma\). - Use differences of nested central areas for a band between two standard-deviation boundaries. - Use symmetry for the one-sided tail.

Solution

1. Part a) is the central two-standard-deviation interval, so its probability is approximately \(95.4\%\). 2. Part b) runs from \(\mu+2\sigma\) to \(\mu+3\sigma\). Its probability is approximately \((99.7\%-95.4\%)/2=2.15\%\). 3. Part c) begins at \(\mu+\sigma\). The right-tail probability is approximately \(50\%-68.3\%/2=15.85\%\).

Answer

a) Approximately \(95.4\%\). b) Approximately \(2.15\%\). c) Approximately \(15.85\%\).

52527512

The lifetime of a certain type of LED bulb is approximately normally distributed with mean \(\mu=15{,}000\,\text{h}\) and standard deviation \(\sigma=800\,\text{h}\). Find the interval \([\mu-k\sigma, \mu+k\sigma]\) associated with each probability. 1. \(68.3\%\) 2. \(95.4\%\) 3. \(99.7\%\)

Hints

- Recall the standard-deviation counts in the \(68\text{-}95\text{-}99.7\) rule. - Match each percentage with its value of \(k\). - Substitute into \([\mu-k\sigma, \mu+k\sigma]\). - Include hours in each interval.

Solution

1. The \(68\text{-}95\text{-}99.7\) rule associates \(68.3\%\) with \(k=1\). Thus, \([15{,}000-800, 15{,}000+800]=[14{,}200, 15{,}800]\,\text{h}\). 2. The probability \(95.4\%\) corresponds to \(k=2\). Thus, \([15{,}000-2\cdot800, 15{,}000+2\cdot800]=[13{,}400, 16{,}600]\,\text{h}\). 3. The probability \(99.7\%\) corresponds to \(k=3\). Thus, \([15{,}000-3\cdot800, 15{,}000+3\cdot800]=[12{,}600, 17{,}400]\,\text{h}\).

Answer

1. \([14{,}200\,\text{h}, 15{,}800\,\text{h}]\) 2. \([13{,}400\,\text{h}, 16{,}600\,\text{h}]\) 3. \([12{,}600\,\text{h}, 17{,}400\,\text{h}]\)

52528812

A manufacturer models the lifetime \(T\), in hours, of an LED bulb as normally distributed with mean \(\mu=1200\) and standard deviation \(\sigma=100\). Use the \(68\text{-}95\text{-}99.7\) rule to estimate each probability. a) A bulb lasts longer than \(1400\) hours. b) A bulb lasts at most \(1100\) hours. c) A bulb lasts between \(1000\) and \(1300\) hours.

Hints

- Express each boundary in standard-deviation units from the mean. - Decide whether each region requires adding areas or taking a complement. - Remember that \(50\%\) of a normal distribution lies on each side of its mean.

Solution

1. Since \(1400=\mu+2\sigma\), the right-tail probability is \((100\%-95.4\%)\div2=2.3\%\). 2. Since \(1100=\mu-\sigma\), the left-tail probability is \((100\%-68.3\%)\div2=15.85\%\). 3. The interval in part c is \([\mu-2\sigma,\mu+\sigma]\). The area from \(\mu-2\sigma\) to \(\mu\) is \(95.4\%\div2=47.7\%\), and the area from \(\mu\) to \(\mu+\sigma\) is \(68.3\%\div2=34.15\%\). 4. Therefore, the probability in part c is \(47.7\%+34.15\%=81.85\%\).

Answer

a) Approximately \(2.3\%\). b) Approximately \(15.85\%\). c) Approximately \(81.85\%\).

52528912

A normal random variable \(X\) has mean \(\mu=100\). Use the \(68\text{-}95\text{-}99.7\) rule to find \(\sigma\) in each independent case. a) \(P(94\le X\le106)\approx0.954\) b) \(P(X\ge109)\approx0.0015\)

Hints

- Match each probability with a one-, two-, or three-standard-deviation region. - Derive a one-sided tail probability by splitting the probability outside a central interval. - Find the distance from each boundary to the mean. - Use symmetry to divide the two-tail probability equally.

Solution

1. a) A central probability of approximately \(0.954\) corresponds to the interval \([\mu-2\sigma, \mu+2\sigma]\). The given interval has radius \(6\), so \(2\sigma=6\) and \(\sigma=3\). 2. b) A one-sided tail probability of approximately \(0.0015\) begins three standard deviations from the mean. Thus, \(100+3\sigma=109\), so \(3\sigma=9\) and \(\sigma=3\).

Answer

a) \(\sigma=3\) b) \(\sigma=3\)

52530512

A beverage company fills bottles with lemonade. The fill amount \(X\) is normally distributed with mean \(\mu=750\,\text{mL}\) and standard deviation \(\sigma=5\,\text{mL}\). Use the Empirical Rule to estimate the probability that a randomly selected bottle contains more than \(760\,\text{mL}\).

Hints

- Determine how many standard deviations the cutoff is above the mean. - Recall the Empirical Rule percentage within two standard deviations. - Use symmetry to divide the probability outside the interval between the two tails.

Solution

1. The cutoff is \(760=750+2\cdot5=\mu+2\sigma\). 2. By the Empirical Rule, about \(95\%\) of values lie between \(\mu-2\sigma\) and \(\mu+2\sigma\). 3. The remaining \(5\%\) is split equally between the two tails because the normal distribution is symmetric. 4. Therefore, \(P(X>760)\approx\frac{0.05}{2}=0.025\).

Answer

Approximately \(0.025\), or \(2.5\%\)

52530612

The length \(X\) of a metal pin is normally distributed with mean \(\mu=120\,\text{mm}\) and standard deviation \(\sigma=0.4\,\text{mm}\). A pin is acceptable when its length is between \(118.8\,\text{mm}\) and \(121.2\,\text{mm}\). Use the Empirical Rule to estimate the percentage of pins that are acceptable.

Hints

- Check whether the endpoints are equally far from the mean. - Express each endpoint as the mean plus or minus a multiple of the standard deviation. - Recall the Empirical Rule percentage within three standard deviations.

Solution

1. The lower endpoint is \(118.8=120-3\cdot0.4=\mu-3\sigma\). 2. The upper endpoint is \(121.2=120+3\cdot0.4=\mu+3\sigma\). 3. By the Empirical Rule, about \(99.7\%\) of values in a normal distribution lie within three standard deviations of the mean.

Answer

Approximately \(99.7\%\) of the pins are acceptable.

52530812

A normal random variable \(X\) has mean \(\mu=50\). The probability that \(X\) lies in \([44,56]\) is approximately \(0.9973\). Find the standard deviation \(\sigma\).

Hints

- Check whether the interval is symmetric about the mean. - Match \(0.9973\) with a familiar central normal probability. - Set the interval radius equal to the corresponding number of standard deviations.

Solution

1. The interval \([44,56]\) is centered at \(50\) and has radius \(6\). 2. A central probability of approximately \(0.9973\) corresponds to an interval extending three standard deviations from the mean. 3. Therefore, \(3\sigma=6\), so \(\sigma=2\).

Answer

\(\sigma=2\)

52533012

A random variable \(X\) is normally distributed with \(\mu=150\) and \(\sigma=25\). Find each probability. Round nonexact decimal probabilities to four decimal places. a) \(P(X\le125)\) b) \(P(150\le X\le200)\) c) \(P(100\le X\le200)\) d) \(P(X>150)\)

Hints

- Check whether each boundary is an exact number of standard deviations from the mean. - Use symmetry for negative standard scores. - Remember that the total area under a density curve is \(1\).

Solution

1. For \(125\), \(z=-1\), so \(P(X\le125)=\Phi(-1)\approx0.1587\). 2. The interval in part b is from \(\mu\) to \(\mu+2\sigma\). Thus, \(P(150\le X\le200)=\Phi(2)-0.5\approx0.4772\). 3. The interval in part c is \([\mu-2\sigma,\mu+2\sigma]\), so \(P(100\le X\le200)=\Phi(2)-\Phi(-2)\approx0.9545\). 4. By symmetry, \(P(X>150)=0.5\).

Answer

a) \(0.1587\). b) \(0.4772\). c) \(0.9545\). d) \(0.5\).

52533712

A continuous random variable \(X\) is normally distributed with \(\mu=80\) and \(\sigma=10\). Find the probability that \(X\) lies in each interval. Round to four decimal places. 1. \(I_1=[60,70]\) 2. \(I_2=[70,80]\) 3. \(I_3=[80,95]\)

Hints

- An interval probability is a difference of cumulative probabilities. - Standardize each endpoint. - Use the standard normal cumulative distribution function. - Use symmetry about the mean to check your results.

Solution

1. For \([60,70]\), the standard scores are \(-2\) and \(-1\). Thus, \(P(60\le X\le70)=\Phi(-1)-\Phi(-2)\approx0.1359\). 2. For \([70,80]\), the standard scores are \(-1\) and \(0\). Thus, \(P(70\le X\le80)=\Phi(0)-\Phi(-1)\approx0.3413\). 3. For \([80,95]\), the standard scores are \(0\) and \(1.5\). Thus, \(P(80\le X\le95)=\Phi(1.5)-\Phi(0)\approx0.4332\).

Answer

1. \(0.1359\). 2. \(0.3413\). 3. \(0.4332\).

52533812

The length \(X\), in millimeters, of a metal pin is normally distributed with \(\mu=100\,\text{mm}\) and \(\sigma=2\,\text{mm}\). Find each probability, rounded to four decimal places. a) The pin is between \(97\,\text{mm}\) and \(103\,\text{mm}\) long. b) The pin is between \(103\,\text{mm}\) and \(105\,\text{mm}\) long. c) The pin is shorter than \(96\,\text{mm}\).

Hints

- Locate each interval relative to the mean. - Standardize each boundary. - A “less than” probability is cumulative area to the left of the boundary. - Use a standard normal table or technology.

Solution

1. The standard scores for part a are \(-1.5\) and \(1.5\). Therefore, \(P(97\le X\le103)=\Phi(1.5)-\Phi(-1.5)\approx0.8664\). 2. The standard scores for part b are \(1.5\) and \(2.5\). Therefore, \(P(103\le X\le105)=\Phi(2.5)-\Phi(1.5)\approx0.0606\). 3. For \(96\), \(z=-2\). Thus, \(P(X<96)=\Phi(-2)\approx0.0228\).

Answer

a) \(0.8664\). b) \(0.0606\). c) \(0.0228\).

52534112

A dairy fills yogurt cups. The fill weight \(X\), in grams, is normally distributed with mean \(\mu=252\) and standard deviation \(\sigma=1.5\). Find each probability, rounded to four decimal places. a) A randomly selected cup contains less than \(250\,\text{g}\). b) A cup contains between \(251\,\text{g}\) and \(254\,\text{g}\). c) A cup contains at least \(255\,\text{g}\).

Hints

- Convert each endpoint to a standard score. - For an interval, subtract the cumulative probability at the lower endpoint from the cumulative probability at the upper endpoint. - For a value above a cutoff, use the upper-tail probability.

Solution

1. For part a, \(z=(250-252)/1.5=-1.3333\ldots\), so \(P(X<250)\approx0.0912\). 2. For part b, the endpoint standard scores are \(-0.6667\ldots\) and \(1.3333\ldots\). Thus, \(P(251\le X\le254)\approx0.6563\). 3. For part c, \(z=(255-252)/1.5=2\), so \(P(X\ge255)=1-\Phi(2)\approx0.0228\).

Answer

a) \(0.0912\). b) \(0.6563\). c) \(0.0228\).

52534212

The lifetime \(X\) of a certain type of LED bulb is approximately normally distributed with mean \(\mu=20{,}000\) hours and standard deviation \(\sigma=800\) hours. Find each probability, rounded to four decimal places. a) A bulb lasts at most \(18{,}800\) hours. b) A bulb lasts between \(19{,}600\) and \(21{,}200\) hours. c) A bulb lasts longer than \(21{,}000\) hours.

Hints

- Convert each endpoint to a standard score. - Use the difference of cumulative probabilities for an interval. - Use the complement of the cumulative probability for an upper tail.

Solution

1. For part a, \(z=(18{,}800-20{,}000)/800=-1.5\), so \(P(X\le18{,}800)=\Phi(-1.5)\approx0.0668\). 2. For part b, the endpoint standard scores are \(-0.5\) and \(1.5\), so \(P(19{,}600\le X\le21{,}200)=\Phi(1.5)-\Phi(-0.5)\approx0.6247\). 3. For part c, \(z=(21{,}000-20{,}000)/800=1.25\), so \(P(X>21{,}000)=1-\Phi(1.25)\approx0.1056\).

Answer

a) \(0.0668\). b) \(0.6247\). c) \(0.1056\).

52534712

A continuous random variable \(X\) is normally distributed with \(\mu=60\) and \(\sigma=12\). Find \(P(X\ge75)\), rounded to four decimal places.

Hints

- Translate “at least” into an inequality. - Standardize the boundary. - Use a complement or symmetry to find a right-tail probability.

Solution

1. Standardize \(75\): \(z=(75-60)/12=1.25\). 2. Use the complement: \(P(X\ge75)=1-\Phi(1.25)\). 3. Therefore, \(P(X\ge75)\approx0.1056\).

Answer

\(P(X\ge75)\approx0.1056\).

52534812

A random variable \(X\) is normally distributed with \(\mu=12\) and \(\sigma=2.5\). Find \(P(10\le X\le15)\), rounded to four decimal places.

Hints

- An interval probability is a difference of cumulative probabilities. - Standardize each endpoint. - Use unrounded cumulative values until the final subtraction.

Solution

1. Standardize the endpoints: \(z_1=(10-12)/2.5=-0.8\) and \(z_2=(15-12)/2.5=1.2\). 2. Therefore, \(P(10\le X\le15)=\Phi(1.2)-\Phi(-0.8)\). 3. Using unrounded cumulative values gives \(P(10\le X\le15)\approx0.6731\).

Answer

\(P(10\le X\le15)\approx0.6731\).

53119712

The diameters of precision components are modeled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). A quality-control sample contains \(400\) independently selected components. a) Find the expected number of components with diameters within one standard deviation of the mean, using the Empirical Rule. b) Find the expected number with diameters outside two standard deviations of the mean. c) Find the expected number with diameters in \([\mu-3\sigma,\mu+3\sigma]\). Give each expected count to one decimal place and also state the corresponding approximate whole-number count.

Hints

- Recall the Empirical Rule probabilities. - Multiply each event probability by \(400\) to obtain an expected count. - Keep the expected value as a decimal even when giving a nearby whole-number interpretation.

Solution

1. The Empirical Rule gives probability approximately \(0.683\) within one standard deviation. Therefore, \(400(0.683)=273.2.\) So the expected count is \(273.2\), corresponding to about \(273\) components. 2. The probability outside two standard deviations is approximately \(1-0.954=0.046.\) Therefore, \(400(0.046)=18.4.\) So the expected count is \(18.4\), corresponding to about \(18\) components. 3. The probability within three standard deviations is approximately \(0.997\). Therefore, \(400(0.997)=398.8.\) So the expected count is \(398.8\), corresponding to about \(399\) components.

Answer

a) Expected count \(273.2\); about \(273\) components. b) Expected count \(18.4\); about \(18\) components. c) Expected count \(398.8\); about \(399\) components.

53275312

The graph shows the probability density function of the fill volume \(X\), in milliliters, of milk bottles. The filled marker shows the peak, and the open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. b) Use the \(68\text{-}95\text{-}99.7\) rule to estimate \(P(495\le X\le505)\). c) Use symmetry and the \(68\text{-}95\text{-}99.7\) rule to estimate \(P(X<490)\).

Figure for problem 532753

Hints

- Read the peak and inflection markers from the graph. - Normal inflection points occur one standard deviation from the mean. - Match one- and two-standard-deviation regions with the Empirical Rule. - Use symmetry for the one-sided tail.

Solution

1. The filled peak is at \(500\), so \(\mu=500\,\text{mL}\). The open inflection markers are at \(495\) and \(505\), so \(\sigma=5\,\text{mL}\). 2. The interval \([495,505]\) is \([\mu-\sigma,\mu+\sigma]\). By the Empirical Rule, its probability is approximately \(68\%\). 3. The cutoff \(490\) is \(\mu-2\sigma\). Approximately \(95\%\) of values lie within two standard deviations, leaving about \(5\%\) outside. 4. By symmetry, half of that outside probability is in the left tail, so \(P(X<490)\approx2.5\%\).

Answer

a) \(\mu=500\,\text{mL}\), \(\sigma=5\,\text{mL}\). b) Approximately \(68\%\). c) Approximately \(2.5\%\).

53275412

The graph shows the density function \(f\) of a normally distributed random variable \(X\). The filled marker shows the peak, and the open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. b) Without calculation, find \(P(X\le3)\). c) Use the Empirical Rule to estimate \(P(2\le X\le4)\).

Figure for problem 532754

Hints

- Read the filled peak and open inflection markers. - Normal inflection points occur at \(\mu\pm\sigma\). - Use symmetry and the fact that the total area is \(1\). - Recall the Empirical-Rule percentage within one standard deviation.

Solution

1. The filled peak is at \(x=3\), so \(\mu=3\). The open inflection markers are at \(x=2\) and \(x=4\), so \(\sigma=1\). 2. By symmetry, half of the total area lies to the left of the mean. Therefore, \(P(X\le3)=0.5\). 3. The interval \([2,4]\) is \([\mu-\sigma,\mu+\sigma]\). By the Empirical Rule, \(P(2\le X\le4)\approx0.68\).

Answer

a) \(\mu=3\), \(\sigma=1\). b) \(0.5\). c) Approximately \(0.68\).

53479512

Two machines fill flour packages with a target weight of \(500\,\text{g}\). The actual fill weights are normally distributed. The graph shows density \(f_1\) for Machine 1 and density \(f_2\) for Machine 2. Filled markers show the peaks, and open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) for each machine. b) Which machine is more precise relative to the target weight? Justify your answer using the graph and the standard deviations.

Figure for problem 534795

Hints

- Read the filled peak markers to identify the means. - Normal inflection points occur at \(\mu\pm\sigma\). - Compare the horizontal distances from each peak to its open markers. - Smaller standard deviation means tighter concentration.

Solution

1. Both filled peak markers occur at \(x=500\), so both machines have mean \(500\,\text{g}\). 2. For \(f_1\), the open inflection markers are at \(498\) and \(502\), so \(\sigma_1=2\,\text{g}\). 3. For \(f_2\), the open inflection markers are at \(495\) and \(505\), so \(\sigma_2=5\,\text{g}\). 4. Machine 1 is more precise because its smaller standard deviation means its fill weights are more tightly concentrated around the target.

Answer

a) Machine 1: \(\mu_1=500\,\text{g}\), \(\sigma_1=2\,\text{g}\). Machine 2: \(\mu_2=500\,\text{g}\), \(\sigma_2=5\,\text{g}\). b) Machine 1 is more precise.

53479912

The graph shows the approximately normal density of freshly baked loaf weights \(X\), in grams. The filled marker shows the peak, and the open markers show the inflection points. a) Determine \(\mu\) and \(\sigma\) from the graph. b) What percentage of loaves weigh more than the mean? c) Use the Empirical Rule to estimate the probability that a loaf weighs between \(980\,\text{g}\) and \(1020\,\text{g}\). d) Estimate the percentage of loaves that weigh less than \(960\,\text{g}\). e) Give the symmetric interval centered at the mean that contains about \(99.7\%\) of loaf weights.

Figure for problem 534799

Hints

- Read the filled peak and open inflection markers from the graph. - Normal inflection points occur one standard deviation from the mean. - Use symmetry of the normal density. - Match one-, two-, and three-standard-deviation regions with the Empirical Rule.

Solution

1. The filled peak marker is at \(1000\), so \(\mu=1000\,\text{g}\). The open markers are at \(980\) and \(1020\), so \(\sigma=20\,\text{g}\). 2. By symmetry, \(50\%\) of loaves weigh more than the mean. 3. The interval \([980,1020]\) is \([\mu-\sigma,\mu+\sigma]\), so its probability is approximately \(0.68\). 4. The cutoff \(960\) is \(\mu-2\sigma\). About \(5\%\) lies outside the central two-standard-deviation interval, so symmetry puts approximately \(2.5\%\) in the lower tail. 5. Three standard deviations equal \(60\,\text{g}\), so the approximate \(99.7\%\) interval is \([940,1060]\).

Answer

a) \(\mu=1000\,\text{g}\), \(\sigma=20\,\text{g}\). b) \(50\%\). c) Approximately \(0.68\). d) Approximately \(2.5\%\). e) \([940\,\text{g},1060\,\text{g}]\).

53480912

The graph shows the density function of a normally distributed random variable \(X\). The filled marker shows the peak. a) Read the mean \(\mu\) from the graph. b) Without calculating normal cumulative probabilities, decide whether \(P(10\le X\le15)\) or \(P(15\le X\le20)\) is greater. Justify your answer using the graph.

Figure for problem 534809

Hints

- Read the location of the peak from the graph. - Compare intervals of equal width. - Probability is area under the density curve, so compare the density heights across the two intervals.

Solution

1. The filled peak is at \(x=10\), so the normal distribution has mean \(\mu=10\). 2. The two intervals have the same width, but every point in \([10,15]\) is closer to the peak than the corresponding point \(5\) units to the right in \([15,20]\). 3. The density is therefore higher over the first interval, so the area under the curve from \(10\) to \(15\) is greater.

Answer

a) \(\mu=10\). b) \(P(10\le X\le15)\) is greater.

53481512

The height \(X\), in inches, of adult women in a region is modeled by a normal distribution with standard deviation \(2.5\) inches. The graph shows the corresponding probability density function, with a filled marker at its peak. a) Read the mean \(\mu\) from the graph and briefly explain why it is located at the graph's peak. b) Use the graph and the Empirical Rule to decide whether \(P(62.5\le X\le67.5)\) is greater than or less than \(P(X<62.5)+P(X>67.5).\) Justify your answer by comparing the corresponding areas.

Figure for problem 534815

Hints

- A normal density is centered at its peak. - Compare the given endpoints with \(\mu\pm\sigma\). - Use the Empirical Rule to compare the central area with the combined tail area.

Solution

1. The filled peak is at \(x=65\). A normal density is symmetric and reaches its maximum at its mean, so \(\mu=65\text{ inches}.\) 2. Since \(\sigma=2.5\), \([62.5,67.5]=[\mu-\sigma,\mu+\sigma].\) 3. By the Empirical Rule, approximately \(68\%\) of the distribution lies in this central interval, leaving about \(32\%\) in the two tails. 4. Therefore, \(P(62.5\le X\le67.5) > P(X<62.5)+P(X>67.5).\)

Answer

a) \(\mu=65\) inches. b) The central probability is greater: approximately \(68\%\) versus \(32\%\) in the two tails.

53483112

The fill weight \(X\), in grams, of coffee packages is normally distributed. The graph shows the cumulative distribution function \(F\). a) Determine the mean \(\mu\) from the graph. b) Estimate the probability that a randomly selected package weighs between \(248\,\text{g}\) and \(252\,\text{g}\). c) Only the heaviest \(10\%\) of packages receive a special label. Estimate the minimum weight required for the label.

Figure for problem 534831

Hints

- The normal CDF equals \(0.5\) at the mean. - Find an interval probability by subtracting two CDF values. - If \(10\%\) are heavier than the cutoff, determine the cumulative proportion below it.

Solution

1. a) For a normal distribution, the mean is the x-value where \(F(x)=0.5\). From the graph, \(\mu\approx250\,\text{g}\). 2. b) Use the difference of cumulative probabilities: \(P(248\le X\le252)=F(252)-F(248)\). 3. From the graph, \(F(252)\approx0.84\) and \(F(248)\approx0.16\). Therefore, the probability is approximately \(0.84-0.16=0.68\). 4. c) If the heaviest \(10\%\) receive the label, then \(90\%\) are at or below the cutoff. We need \(F(x)=0.90\). 5. Reading the graph gives a cutoff of about \(253\,\text{g}\).

Answer

a) \(\mu\approx250\,\text{g}\) b) Approximately \(0.68\), or \(68\%\) c) Approximately \(253\,\text{g}\)

55630012

Two normal probability plots are shown. In such a plot, data that are reasonably consistent with a normal model tend to follow a straight reference line. a) Which panel, 1 or 2, gives stronger evidence that a normal model is reasonable? b) What feature of the other panel argues against a normal model?

Figure for problem 556300

Hints

- Focus on the overall pattern of the points, not one isolated point. - Ask which panel stays close to a straight line from the lower tail through the upper tail. - Systematic curvature is different from small random scatter around a line.

Solution

1. In panel 1, the plotted points stay close to the straight reference line across the full range. 2. Therefore, panel 1 is more consistent with a normal model. 3. In panel 2, the points show systematic curvature: the lower points lie above the reference line while the upper tail bends well above it. 4. A systematic curved pattern rather than random small departures from a line is evidence against a normal model.

Answer

a) Panel 1. b) Panel 2 shows systematic curvature away from the straight reference line, especially in the upper tail.

52511012

Anna studies a normal random variable \(X\sim N(\mu,\sigma^2)\) with density function \(\phi\). Anna notices that \(\phi(\mu)=\frac{1}{\sigma\sqrt{2\pi}}>0\) and concludes that the probability of obtaining exactly the mean, \(P(X=\mu)\), must be positive. Evaluate Anna's conclusion. Explain the fundamental difference between the density value \(\phi(x)\) and the probability \(P(X=x)\) for a continuous distribution.

Hints

- How are probabilities represented graphically for a continuous distribution? - Can a probability exceed \(1\)? Can a density value exceed \(1\)? - Distinguish the height of a curve from an area under the curve.

Solution

1. The conclusion is false. For a continuous random variable, the probability of every exact value is \(0\). 2. The value \(\phi(x)\) is a probability density, not a probability. It describes the local concentration of probability, and a density value may even be greater than \(1\). 3. Probabilities are areas under the density curve. A single point has width \(0\), so the area above that point is \(0\), regardless of the curve's height. Therefore, \(P(X=\mu)=0\).

Answer

The conclusion is false. The value \(\phi(x)\) is a local density, while probability is area under the density curve. A single point has zero width, so \(P(X=\mu)=0\) even though the density is greatest at \(x=\mu\).

52511212

The weight \(X\) of a coffee package is normally distributed with mean \(\mu=500\,\text{g}\) and standard deviation \(\sigma=4\,\text{g}\). 1. Explain why the probability that a package weighs exactly \(500\,\text{g}\) is \(0\). 2. Package weights are recorded to the nearest gram. Find the probability that a package is recorded as \(500\,\text{g}\). Round to four decimal places. 3. Find the probability that a package weight rounded to the nearest tenth of a gram is recorded as \(500.0\,\text{g}\). Round to four decimal places and compare this probability with your result from part 2.

Hints

- For a continuous variable, ask how much probability belongs to one exact point. - Which weights round to \(500\) when rounding to the nearest gram? - How does that interval change when rounding to the nearest tenth? - Standardize the interval endpoints and use the standard normal cumulative distribution function \(\Phi\).

Solution

1. A normal random variable is continuous, so probability is assigned to intervals rather than to individual points. A single exact value has zero width, so \(P(X=500)=0\). 2. A weight rounds to \(500\,\text{g}\) when \(499.5\le X<500.5.\) Standardizing gives \(P(499.5\le X<500.5) =\Phi(0.125)-\Phi(-0.125) \approx0.0995.\) 3. A weight rounds to \(500.0\,\text{g}\) when \(499.95\le X<500.05.\) Thus, \(P(499.95\le X<500.05) =\Phi(0.0125)-\Phi(-0.0125) \approx0.0100.\) The second rounding interval is one-tenth as wide as the first, so its probability is much smaller.

Answer

1. \(P(X=500)=0\). 2. \(P(499.5\le X<500.5)\approx0.0995\). 3. \(P(499.95\le X<500.05)\approx0.0100\). This is much smaller because the rounding interval is narrower.

52512812

The fill weight \(X\), in grams, of flour packages from a filling machine is approximately normally distributed with \(\mu=505\) and \(\sigma=2\). Find each probability. Round nonexact decimal probabilities to four decimal places. a) A package contains at most \(505\,\text{g}\). b) The fill weight is between \(501\,\text{g}\) and \(509\,\text{g}\). c) A package contains less than the labeled weight of \(500\,\text{g}\). d) The fill weight is exactly \(505.0\,\text{g}\). e) The fill weight differs from the mean by more than \(5\,\text{g}\).

Hints

- Express each value as a number of standard deviations from the mean. - Use symmetry to simplify calculations. - For a deviation greater than a given amount, account for both tails. - A continuous random variable assigns probability \(0\) to one exact value.

Solution

1. Since \(505\) is the mean, \(P(X\le505)=0.5.\) 2. The interval \([501{,}509]\) is \([\mu-2\sigma,\mu+2\sigma]\), so \(P(501\le X\le509)=\Phi(2)-\Phi(-2)\approx0.9545.\) 3. For \(500\), \(z=\frac{500-505}{2}=-2.5,\) so \(P(X<500)=\Phi(-2.5)\approx0.0062.\) 4. Because \(X\) is continuous, \(P(X=505.0)=0.\) 5. \(P(|X-505|>5)=2\Phi(-2.5)\approx0.0124.\)

Answer

a) \(0.5\) b) \(0.9545\) c) \(0.0062\) d) \(0\) e) \(0.0124\)

52513812

IQ scores in a population are approximately normally distributed with mean \(\mu=100\) and standard deviation \(\sigma=15\). Use a standard normal table or technology to find each probability. Round probabilities to four decimal places; equivalent percentages may be given to two decimal places. 1. A randomly selected person has an IQ score between \(85\) and \(115\). 2. A person has an IQ score of at most \(70\). 3. A person has an IQ score greater than \(130\). 4. A person has an IQ score between \(110\) and \(120\).

Hints

- Standardize each value before using the standard normal cumulative distribution function. - Use symmetry for negative standard scores and matching tails. - The empirical rule can help check whether results near one or two standard deviations from the mean are reasonable.

Solution

1. The standard scores are \(-1\) and \(1\), so \(P(85\le X\le115)=\Phi(1)-\Phi(-1)\approx0.6827.\) 2. For \(70\), \(z=-2\), so \(P(X\le70)=\Phi(-2)\approx0.0228.\) 3. For \(130\), \(z=2\), so \(P(X>130)=1-\Phi(2)\approx0.0228.\) 4. The standard scores are \(2/3\) and \(4/3\), so \(P(110\le X\le120) =\Phi\left(\frac43\right)-\Phi\left(\frac23\right) \approx0.1613.\)

Answer

1. \(0.6827\), or \(68.27\%\) 2. \(0.0228\), or \(2.28\%\) 3. \(0.0228\), or \(2.28\%\) 4. \(0.1613\), or \(16.13\%\)

52514512

The fill volume \(X\), in milliliters, of bottled water is modeled by a normal distribution with mean \(\mu=500\,\text{mL}\) and standard deviation \(\sigma=3\,\text{mL}\). a) Find the probability that a randomly selected bottle contains between \(497\,\text{mL}\) and \(506\,\text{mL}\). Round to four decimal places. b) A bottle is considered underfilled if it contains less than \(495\,\text{mL}\). Find the percentage of bottles that are underfilled, rounded to two decimal places. c) For a continuous random variable, \(P(X=500)=0\). In practice, measurements are rounded to the nearest whole milliliter. Find the probability that a bottle is recorded as containing exactly \(500\,\text{mL}\). Give the probability to four decimal places and the equivalent percentage to two decimal places.

Hints

- Standardize each boundary. - Use symmetry of the standard normal cumulative distribution function for negative values. - Translate rounding to the nearest whole number into an interval of original measurements. - Distinguish an exact continuous value from a recorded rounded value.

Solution

1. Standardizing gives \(P(497\le X\le506)=P(-1\le Z\le2) =\Phi(2)-\Phi(-1) \approx0.8186.\) 2. For \(495\), \(z=\frac{495-500}{3}=-\frac53,\) so \(P(X<495)=\Phi\left(-\frac53\right)\approx0.0478,\) or \(4.78\%\). 3. Rounding to \(500\,\text{mL}\) corresponds to \(499.5\le X<500.5.\) Therefore, \(P(499.5\le X<500.5) =\Phi\left(\frac16\right)-\Phi\left(-\frac16\right) \approx0.1324,\) or \(13.24\%\).

Answer

a) \(0.8186\) b) \(4.78\%\) c) \(0.1324\), or \(13.24\%\)

52514612

The standard normal density is \(\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}.\) a) Show by differentiation that \(g(z)=-e^{-z^2/2}\) is an antiderivative of \(h(z)=ze^{-z^2/2}.\) b) Use part a) to evaluate \(\int_{-\infty}^{\infty} z\phi(z)\,\mathrm{d}z.\) c) Explain why the value of this improper integral is the mean of the standard normal distribution, and relate the result to the symmetry of the density.

Hints

- Apply the chain rule to the exponential function. - Determine the limit of \(e^{-z^2/2}\) as \(z\to\pm\infty\). - Interpret \(\int x f(x)\,\mathrm{d}x\) as an accumulated first moment. - Use odd-function symmetry as a check on the integral.

Solution

1. By the chain rule, \(\frac{\mathrm{d}}{\mathrm{d}z}\left(-e^{-z^2/2}\right) =ze^{-z^2/2}.\) Thus, \(g\) is an antiderivative of \(h\). 2. \(\int_{-\infty}^{\infty}z\phi(z)\,\mathrm{d}z =\frac{1}{\sqrt{2\pi}} \left[-e^{-z^2/2}\right]_{-\infty}^{\infty} =0,\) because \(e^{-z^2/2}\to0\) as \(z\to\pm\infty\). 3. For a continuous random variable with density \(f\), the mean is the accumulation \(E(X)=\int_{-\infty}^{\infty}x f(x)\,\mathrm{d}x.\) Therefore, the integral in part b) is the mean of the standard normal distribution. 4. The result \(0\) is also consistent with symmetry: \(z\phi(z)\) is an odd function, so the negative and positive contributions cancel over symmetric limits.

Answer

a) \(\frac{\mathrm{d}}{\mathrm{d}z}(-e^{-z^2/2})=ze^{-z^2/2}\). b) \(0\). c) The integral is \(E(Z)\), so the standard normal mean is \(0\). The result agrees with symmetry because \(z\phi(z)\) is odd.

52514812

The standard normal density is symmetric about the \(y\)-axis, and the total area under the graph is \(1\). a) Use symmetry and total probability to explain why \(P(Z\le0)=0.5\). b) Given that \(P(-1\le Z\le1)\approx0.6827\), use symmetry to find \(P(Z\ge1)\). Round to four decimal places. c) Describe how the density curve of a normal distribution changes when \(\sigma\) increases while \(\mu\) remains fixed.

Hints

- Use the mirror symmetry of the normal curve. - After subtracting the middle area from \(1\), divide the remaining area between the two tails. - Relate greater standard deviation to greater horizontal spread while total area stays fixed.

Solution

1. Symmetry about \(z=0\) makes the area to the left of \(0\) equal to the area to the right. Since the total area is \(1\), each half has area \(0.5\). Thus, \(P(Z\le0)=0.5.\) 2. The area outside \([-1,1]\) is approximately \(1-0.6827=0.3173.\) Symmetry divides this equally between the two tails, so \(P(Z\ge1)\approx\frac{0.3173}{2}=0.15865\approx0.1587.\) 3. A larger standard deviation means greater spread. The normal curve becomes wider and flatter, while remaining symmetric about the same mean and keeping total area \(1\).

Answer

a) \(P(Z\le0)=0.5\). b) \(P(Z\ge1)\approx0.1587\). c) The curve becomes wider and flatter, while remaining centered at the same mean.

52519112

A filling line checks the masses of \(100\) flour packages, rounded to the nearest gram. The results are shown below. <table> <tr> <td>Mass (\(\text{g}\))</td> <td>497</td><td>498</td><td>499</td><td>500</td><td>501</td><td>502</td><td>503</td> </tr> <tr> <td>Number of packages</td> <td>3</td><td>12</td><td>25</td><td>30</td><td>18</td><td>9</td><td>3</td> </tr> </table> a) Describe the shape, center, and spread of the observed distribution qualitatively. b) Find the sample mean \(\overline{x}\). c) Find the signed percent deviation of \(\overline{x}\) from the target mass of \(500\,\text{g}\), using the target mass as the reference value.

Hints

- Describe the overall pattern before calculating. - Use a weighted mean because values occur with different frequencies. - Divide the signed difference by the target value for the percent deviation.

Solution

1. The distribution is unimodal and approximately symmetric, with its center near \(500\,\text{g}\) and relatively small spread. 2. The weighted total is \(497(3)+498(12)+499(25)+500(30)+501(18)+502(9)+503(3)=49{,}987.\) Therefore, \(\overline{x}=\frac{49{,}987}{100}=499.87\,\text{g}.\) 3. The signed percent deviation is \(\frac{499.87-500}{500}\cdot100\%=-0.026\%.\) The sample mean is \(0.026\%\) below the target.

Answer

a) Unimodal, approximately symmetric, centered near \(500\,\text{g}\), with small spread. b) \(\overline{x}=499.87\,\text{g}\). c) \(-0.026\%\), meaning \(0.026\%\) below the target.

52519212

A factory measures the lengths of \(200\) metal pins in millimeters. The sample results are shown below. <table> <tr> <td>Length (\(\text{mm}\))</td> <td>\(24.8\)</td><td>\(24.9\)</td><td>\(25.0\)</td><td>\(25.1\)</td><td>\(25.2\)</td> </tr> <tr> <td>Count</td> <td>12</td><td>45</td><td>82</td><td>48</td><td>13</td> </tr> </table> a) Describe the shape and center of the observed distribution qualitatively. b) Find the empirical mean \(\mu_{\text{emp}}\) and empirical standard deviation \(s_{\text{emp}}\), using division by \(n\) for the empirical variance. Round \(s_{\text{emp}}\) to four decimal places. c) Find the signed percent deviation of \(\mu_{\text{emp}}\) from \(25.0\,\text{mm}\), using \(25.0\,\text{mm}\) as the reference value.

Hints

- Look for a central peak and roughly symmetric decrease. - Organize weighted calculations for the mean and standard deviation. - Standard deviation measures spread about the mean. - Divide the signed difference by the reference value for percent deviation.

Solution

1. The frequencies peak near \(25.0\,\text{mm}\) and decrease approximately symmetrically toward both ends. 2. The weighted total is \(24.8(12)+24.9(45)+25.0(82)+25.1(48)+25.2(13)=5000.5.\) Thus, \(\mu_{\text{emp}}=\frac{5000.5}{200}=25.0025\,\text{mm}.\) 3. The empirical variance is \(\frac1{200}\sum f_i(x_i-25.0025)^2 \approx0.00964375\,\text{mm}^2,\) so \(s_{\text{emp}}\approx0.0982\,\text{mm}.\) 4. The signed percent deviation is \(\frac{25.0025-25.0}{25.0}\cdot100\%=0.01\%.\)

Answer

a) Approximately symmetric and centered near \(25.0\,\text{mm}\). b) \(\mu_{\text{emp}}=25.0025\,\text{mm}\) and \(s_{\text{emp}}\approx0.0982\,\text{mm}\). c) \(0.01\%\) above the reference value.

52519612

Vehicle speeds \(X\), in miles per hour, on a highway segment are modeled by the normal density \(g(x)=\frac{1}{\sqrt{18\pi}}e^{-\frac{(x-70)^2}{18}}.\) a) State the mean \(\mu\) and standard deviation \(\sigma\). b) Find the probability that a randomly recorded vehicle is traveling faster than \(76\) miles per hour. Give a standard-normal expression and a decimal approximation to four decimal places.

Hints

- Compare the given function with the standard normal-density form. - Remember that the exponent denominator is \(2\sigma^2\). - Translate “faster than” into a right-tail probability. - Standardize the speed boundary.

Solution

1. Compare the density with \(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}.\) 2. The center is \(\mu=70\), and \(2\sigma^2=18\), so \(\sigma=3\,\text{mph}.\) 3. For \(76\,\text{mph}\), \(z=\frac{76-70}{3}=2.\) 4. Therefore, \(P(X>76)=1-\Phi(2)\approx0.0228.\)

Answer

a) \(\mu=70\,\text{mph}\) and \(\sigma=3\,\text{mph}\). b) \(1-\Phi(2)\approx0.0228\), or \(2.28\%\).

52520112

The fill weight \(X\), in grams, of coffee-bean packages is normally distributed with mean \(\mu=250\,\text{g}\) and standard deviation \(\sigma=4\,\text{g}\). Find each probability. Round nonzero nonexact probabilities to four decimal places. a) A package contains less than \(245\,\text{g}\). b) A package contains between \(248\,\text{g}\) and \(255\,\text{g}\). c) A package's fill weight differs from the mean by more than \(6\,\text{g}\). d) A package contains exactly \(250\,\text{g}\).

Hints

- Standardize each boundary. - Use a CDF difference for the bounded interval. - Use both tails for the absolute-deviation event. - Recall the probability of one exact value for a continuous random variable.

Solution

1. For \(245\), \(z=-1.25\), so \(P(X<245)\approx0.1056\). 2. For \(248\) and \(255\), the standard scores are \(-0.5\) and \(1.25\), so \(P(248\le X\le255)\approx0.5858\). 3. A deviation greater than \(6\,\text{g}\) means \(|Z|>1.5\). Thus, \(P(|X-250|>6)=2[1-\Phi(1.5)]\approx0.1336\). 4. Because \(X\) is continuous, \(P(X=250)=0\).

Answer

a) \(0.1056\). b) \(0.5858\). c) \(0.1336\). d) \(0\).

52520212

The mass \(M\), in grams, of a variety of apple is normally distributed with mean \(\mu=150\,\text{g}\) and standard deviation \(\sigma=12\,\text{g}\). Find each probability. Round nonzero nonexact probabilities to four decimal places. a) An apple weighs at most \(140\,\text{g}\). b) An apple weighs at least \(165\,\text{g}\). c) An apple weighs between \(145\,\text{g}\) and \(155\,\text{g}\). d) An apple weighs exactly \(150.0\,\text{g}\).

Hints

- Translate “at most” and “at least” to the correct normal tail areas. - Use symmetry for the interval centered at the mean. - Preserve exact standard scores until the final CDF evaluation. - A continuous random variable has zero probability at one exact value.

Solution

1. For \(140\), \(z=-5/6\), so \(P(M\le140)\approx0.2023\). 2. For \(165\), \(z=1.25\), so \(P(M\ge165)=1-\Phi(1.25)\approx0.1056\). 3. The endpoints \(145\) and \(155\) have standard scores \(-5/12\) and \(5/12\), so \(P(145\le M\le155)\approx0.3231\). 4. Since \(M\) is continuous, \(P(M=150.0)=0\).

Answer

a) \(0.2023\). b) \(0.1056\). c) \(0.3231\). d) \(0\).

52520612

For a normally distributed random variable \(X\) with mean \(\mu\), the events \(X<145\) and \(X>155\) have equal probabilities. a) Find \(\mu\). b) Give an interval with the same probability as \((150, 153)\). c) Find an interval of the form \((-\infty, k)\) that has the same probability as \((158, \infty)\).

Hints

- Equal tail areas are located symmetrically about the mean. - Use the midpoint of the two tail boundaries. - Reflect later intervals across the mean found in part a.

Solution

1. a) Equal tail probabilities occur at equal distances from the mean. Thus, \(\mu=\frac{145+155}{2}=150\). 2. b) Reflecting \((150, 153)\) across \(150\) gives \((147, 150)\). 3. c) Since \(158\) is \(8\) units above the mean, the reflected boundary is \(150-8=142\). Therefore, the equal-probability interval is \((-\infty, 142)\).

Answer

a) \(\mu=150\) b) \((147, 150)\) c) \((-\infty, 142)\)

52521112

A normally distributed random variable \(X\) has \(\mu=40\) and \(\sigma=8\). To find \(P(32\le X\le56)\), Emre proposes \(P(X\le56)-P(X\le40)\). a) Explain Emre's error. b) Find the correct value of \(P(32\le X\le56)\), rounded to four decimal places.

Hints

- Write a general interval probability \(P(a\le X\le b)\) using cumulative probabilities. - Compare the requested endpoints with those in the proposed expression. - Standardize both correct endpoints.

Solution

1. Subtracting \(P(X\le40)\) gives only the probability over \([40,56]\). The lower boundary of the requested interval is \(32\), not \(40\). 2. The correct setup is \(P(X\le56)-P(X\le32)\). 3. Standardizing gives \(z_{56}=2\) and \(z_{32}=-1\). 4. Therefore, \(P(32\le X\le56)=\Phi(2)-\Phi(-1)\approx0.8186\).

Answer

a) Emre subtracts the cumulative probability at the mean instead of at the lower endpoint \(32\), so the proposed expression finds only \(P(40\le X\le56)\). b) \(P(32\le X\le56)\approx0.8186\).

52521212

A random variable \(X\) is normally distributed with \(\mu=10\) and \(\sigma=2\). To find \(P(7\le X\le13)\), Camila calculates \(\Phi(13-10)-\Phi(7-10)\). a) Explain why this setup is incorrect, and correct it. b) Find \(P(7\le X\le13)\), rounded to four decimal places.

Hints

- Recall the full standardization formula. - Identify the role of the standard deviation in a standard score. - Use symmetry about the mean after standardizing.

Solution

1. Camila subtracts the mean but does not divide by the standard deviation. 2. The correct setup is \(\Phi\left(\frac{13-10}{2}\right)-\Phi\left(\frac{7-10}{2}\right)=\Phi(1.5)-\Phi(-1.5)\). 3. By symmetry, \(\Phi(1.5)-\Phi(-1.5)=2\Phi(1.5)-1\approx0.8664\).

Answer

a) The division by \(\sigma=2\) is missing. The correct expression is \(\Phi\left(\frac{13-10}{2}\right)-\Phi\left(\frac{7-10}{2}\right)\). b) \(P(7\le X\le13)\approx0.8664\).

52521312

A random variable \(X\) is normally distributed with mean \(\mu=45\) and standard deviation \(\sigma=8\). a) Find \(P(37\le X\le53)\), rounded to four decimal places. b) The standard deviation remains \(8\), but the mean changes to \(40\). Without recalculating the probability, explain whether the probability over \([37,53]\) becomes larger or smaller.

Hints

- Express each endpoint in standard-deviation units from the mean. - Compare the interval midpoint with the two possible means. - Think about where a normal density is highest.

Solution

1. The endpoints are one standard deviation below and above the mean, so \(P(37\le X\le53) =\Phi(1)-\Phi(-1) \approx0.6827.\) 2. The midpoint of \([37,53]\) is \(45\). For a fixed-width interval, a normal distribution places the greatest area in the interval when its mean is at the midpoint. 3. Moving the mean from \(45\) to \(40\) shifts the highest-density part of the curve away from the interval center, so the probability becomes smaller.

Answer

a) \(P(37\le X\le53)\approx0.6827\). b) The probability becomes smaller.

52521412

A random variable \(X\) is normally distributed with \(\mu_X=10\) and \(\sigma_X=4\). a) Find \(P(8\le X\le12)\), rounded to four decimal places. b) Another random variable \(Y\) is normally distributed with \(\mu_Y=10\) and \(\sigma_Y=8\). Find \(P(8\le Y\le12)\), rounded to four decimal places, and explain the difference using the shapes of the density curves.

Hints

- Standardize the endpoints for each distribution. - Compare what the same fixed interval represents in standard-deviation units. - Relate larger standard deviation to a wider, flatter density curve.

Solution

1. For \(X\), the standard scores are \(-0.5\) and \(0.5\), so \(P(8\le X\le12)=\Phi(0.5)-\Phi(-0.5)\approx0.3829\). 2. For \(Y\), the standard scores are \(-0.25\) and \(0.25\), so \(P(8\le Y\le12)=\Phi(0.25)-\Phi(-0.25)\approx0.1974\). 3. The probability is smaller for \(Y\). Its larger standard deviation produces a wider, lower density curve, so less probability lies in the same fixed interval around the mean.

Answer

a) \(P(8\le X\le12)\approx0.3829\). b) \(P(8\le Y\le12)\approx0.1974\), which is smaller because \(Y\) has greater spread.

52523512

The fill weight \(X\), in grams, of sugar packages is normally distributed with mean \(\mu=500\) and standard deviation \(\sigma=10\). a) Find the probability that a randomly selected package has a fill weight of at most \(485\,\text{g}\). Round to four decimal places. b) A supermarket receives \(50\) independent packages. Using the unrounded probability from part a), find the probability that at most \(3\) have fill weights of at most \(485\,\text{g}\). Round to four decimal places. c) Evaluate \(F(510)-F(490)\), where \(F\) is the cumulative distribution function of \(X\), and interpret the result in context. Round to four decimal places.

Hints

- Standardize the package-weight cutoff first. - In part b), use the normal probability from part a) as the Bernoulli success probability without rounding it first. - Interpret a difference of CDF values as an interval probability.

Solution

1. For \(485\), the standard score is \(-1.5\), so \(p=P(X\le485)=\Phi(-1.5)\approx0.0668\). 2. Let \(K\) be the number of the fifty packages with fill weight at most \(485\,\text{g}\). Then \(K\sim\operatorname{Bin}(50,p)\), using the unrounded value \(p=\Phi(-1.5)\). Thus, \(P(K\le3)\approx0.5689\). 3. \(F(510)-F(490)=P(490<X\le510)\). The standard scores are \(-1\) and \(1\), so the probability is approximately \(0.6827\).

Answer

a) \(0.0668\). b) \(0.5689\). c) \(0.6827\); it is the probability that a randomly selected package has fill weight between \(490\,\text{g}\) and \(510\,\text{g}\).

52523612

The lifetime \(X\), in hours, of an electronic component is normally distributed with mean \(\mu=2000\) and standard deviation \(\sigma=250\). a) Find the probability that a component lasts longer than \(2400\) hours. Round to four decimal places. b) A device contains \(20\) independent components of this type. Using the unrounded probability from part a), find the probability that at least one lasts longer than \(2400\) hours. Round to four decimal places. c) Interpret and evaluate \(F(2200)-F(1800)\), where \(F\) is the cumulative distribution function of \(X\). Round to four decimal places.

Hints

- Standardize the lifetime threshold in part a). - Use the complement of no component exceeding the threshold in part b). - Keep the normal tail probability unrounded until the final binomial-style calculation. - Interpret a CDF difference as an interval probability.

Solution

1. For \(2400\), the standard score is \(1.6\), so \(p=P(X>2400)=1-\Phi(1.6)\approx0.0548\). 2. For twenty independent components, the probability that at least one exceeds \(2400\) hours is \(1-(1-p)^{20}\approx0.6760\), using the unrounded value of \(p\). 3. \(F(2200)-F(1800)=P(1800<X\le2200)\). The standard scores are \(-0.8\) and \(0.8\), so the probability is approximately \(0.5763\).

Answer

a) \(0.0548\). b) \(0.6760\). c) \(0.5763\); it is the probability that a component lasts between \(1800\) and \(2200\) hours.

52524812

The diameter \(X\), in millimeters, of precision steel balls is normally distributed with mean \(\mu=10\) and standard deviation \(\sigma=0.02\). a) Find the probabilities that a ball's diameter is in \([9.96,10.04]\), in \([9.99,10.01]\), and exactly \(10.000\,\text{mm}\). Round nonzero interval probabilities to five decimal places. b) A diameter recorded to the nearest hundredth of a millimeter is reported as \(10.00\,\text{mm}\) when \(9.995\le X<10.005\). Find this probability to five decimal places and explain its meaning. c) Machine wear increases the standard deviation to \(0.04\,\text{mm}\) while the mean remains \(10\,\text{mm}\). Find the percent decrease in the probability of the tolerance interval \([9.96,10.04]\), rounded to two decimal places.

Hints

- Convert tolerance endpoints to standard scores. - Distinguish an exact continuous value from a rounded recorded value. - Translate the recording rule into an interval before finding its probability. - Compare the old and new tolerance probabilities using percent decrease.

Solution

1. For \([9.96,10.04]\), the standard scores are \(-2\) and \(2\), so \(P(9.96\le X\le10.04) =\Phi(2)-\Phi(-2) \approx0.95450.\) 2. For \([9.99,10.01]\), the standard scores are \(-0.5\) and \(0.5\), so \(P(9.99\le X\le10.01) \approx0.38292.\) 3. Because \(X\) is continuous, \(P(X=10.000)=0.\) 4. A recorded value of \(10.00\,\text{mm}\) corresponds to \(9.995\le X<10.005.\) The standard scores are \(-0.25\) and \(0.25\), so \(P(9.995\le X<10.005) \approx0.19741.\) This is the probability that the measured diameter rounds to \(10.00\,\text{mm}\). 5. With \(\sigma=0.04\), the tolerance endpoints are one standard deviation from the mean, so the new tolerance probability is \(\Phi(1)-\Phi(-1)\approx0.68269.\) 6. The percent decrease is \(\frac{0.95450-0.68269}{0.95450}\cdot100\% \approx28.48\%.\)

Answer

a) \(0.95450,\ 0.38292,\ 0\), respectively. b) \(0.19741\); this is the probability that the diameter rounds to \(10.00\,\text{mm}\). c) Approximately \(28.48\%\).

52526312

A normal random variable \(X\) has mean \(\mu=500\) and standard deviation \(\sigma=50\). Use the \(68\text{-}95\text{-}99.7\) rule to estimate each probability. a) \(P(450\le X\le550)\) b) \(P(400\le X\le500)\) c) \(P(450\le X\le650)\)

Hints

- Express each endpoint as a number of standard deviations from the mean. - Use symmetry about the mean. - Split an asymmetric interval at the mean. - Apply the percentages from the \(68\text{-}95\text{-}99.7\) rule.

Solution

1. a) The interval is \([\mu-\sigma, \mu+\sigma]\), so the probability is approximately \(68.3\%\). 2. b) The interval is from \(\mu-2\sigma\) to \(\mu\). By symmetry, it contains half of the central \(95.4\%\): \(95.4\%\div2=47.7\%\). 3. c) The interval runs from \(\mu-\sigma\) to \(\mu+3\sigma\). The area from \(\mu-\sigma\) to \(\mu\) is \(68.3\%\div2=34.15\%\), and the area from \(\mu\) to \(\mu+3\sigma\) is \(99.7\%\div2=49.85\%\). 4. Adding gives \(34.15\%+49.85\%=84.0\%\).

Answer

a) Approximately \(68.3\%\) b) Approximately \(47.7\%\) c) Approximately \(84.0\%\)

52526512

The fill weight \(X\), in grams, of flour packages is normally distributed with mean \(\mu=504\) and standard deviation \(\sigma=3\). The labeled fill weight is \(500\,\text{g}\). a) Find the probability that a randomly selected package contains between \(501\,\text{g}\) and \(507\,\text{g}\). Round to four decimal places. b) The manufacturer wants no more than \(3\%\) of packages to contain less than \(495\,\text{g}\), while keeping \(\sigma=3\,\text{g}\). Use an inverse normal calculation or systematic trial to find the smallest mean, to the nearest tenth of a gram, that meets the requirement.

Hints

- Standardize the endpoints in part a). - Determine how increasing the mean changes the lower-tail probability. - Relate the required tail probability to a standard normal quantile. - Test nearby tenths of a gram if using systematic trial.

Solution

1. The endpoints in part a) have standard scores \(-1\) and \(1\), so \(P(501\le X\le507)=\Phi(1)-\Phi(-1)\approx0.6827.\) 2. For part b), require \(P(X<495)\le0.03.\) The \(0.03\) standard-normal quantile is approximately \(-1.8808\), so \(\frac{495-\mu}{3}\le-1.8808.\) Therefore, \(\mu\ge500.6424.\) 3. The smallest mean to the nearest tenth that satisfies the requirement is \(\mu=500.7\,\text{g}.\) A mean of \(500.6\,\text{g}\) gives a lower-tail probability slightly above \(0.03\), while \(500.7\,\text{g}\) gives one below \(0.03\).

Answer

a) \(0.6827\). b) \(\mu=500.7\,\text{g}\).

52527112

A normal random variable \(X\) has mean \(\mu=450\) and standard deviation \(\sigma=40\). Use normal-distribution rules or critical values to answer each question. a) Find an interval symmetric about \(\mu\) that contains approximately \(90\%\) of the values. Round the interval endpoints to the nearest tenth. b) Find an interval symmetric about \(\mu\) such that \(P(X\in I)\approx0.954\). c) Estimate \(P(X\le330)\) using the \(68\text{-}95\text{-}99.7\) rule.

Hints

- Match each central probability with a standard-normal critical value or empirical-rule distance. - Use symmetry about the mean. - For a one-sided tail beyond \(3\sigma\), split the outside probability equally.

Solution

1. A central \(90\%\) interval uses \(z^*\approx1.645\). Thus, \(450\pm1.645(40)=450\pm65.8,\) so \(I\approx[384.2{,}515.8].\) 2. A central probability of approximately \(0.954\) corresponds to two standard deviations from the mean: \(I=[450-2(40),450+2(40)]=[370{,}530].\) 3. Since \(330=450-3(40)=\mu-3\sigma,\) the total probability outside the central three-standard-deviation interval is approximately \(1-0.997=0.003.\) By symmetry, the left tail is approximately \(0.003/2=0.0015.\)

Answer

a) \(I\approx[384.2{,}515.8]\). b) \(I=[370{,}530]\). c) \(P(X\le330)\approx0.0015\).

52527212

A normal random variable \(X\) has mean \(\mu=1200\) and standard deviation \(\sigma=150\). a) Using \(z^*=2.58\) for a central \(99\%\) interval, give the interval symmetric about the mean that contains approximately \(99\%\) of the values. b) Use the \(68\text{-}95\text{-}99.7\) rule to estimate \(P(X>1650)\). c) Use the same rule to find \(c\) such that \(P(X\le c)\approx0.977\).

Hints

- Use the supplied critical value only in part a). - Split the outside probability equally between two symmetric tails. - Relate a cumulative probability of about \(0.977\) to the upper endpoint of the central two-standard-deviation interval.

Solution

1. For part a), \(1200\pm2.58(150)=1200\pm387\), giving \([813,1587]\). 2. Since \(1650=\mu+3\sigma\), the total probability outside the central three-standard-deviation interval is about \(0.003\). The right tail is about \(0.0015\). 3. A cumulative probability of about \(0.977\) occurs at \(\mu+2\sigma\), because \(0.5+0.954/2=0.977\). Therefore, \(c=1500\).

Answer

a) \([813,1587]\). b) Approximately \(0.0015\). c) \(c=1500\).

52528112

A normal random variable \(X\) has mean \(\mu=60\). In each independent case, use normal-distribution rules to find the standard deviation \(\sigma\). a) \(P(X\le72)\approx0.841\) b) \(P(54\le X\le66)\approx0.954\)

Hints

- Use the fact that half of a normal distribution lies below the mean. - Match each probability with a standard-deviation boundary. - Locate the given region on the bell curve. - Distinguish between a one-sided cumulative probability and a central interval.

Solution

1. a) For a normal distribution, \(P(X\le\mu+\sigma)\approx0.8415\). Therefore, \(72\) is approximately one standard deviation above the mean, so \(60+\sigma=72\) and \(\sigma=12\). 2. b) A central probability of approximately \(0.954\) corresponds to the interval \([\mu-2\sigma, \mu+2\sigma]\). The interval \([54, 66]\) is centered at \(60\) and has half-width \(6\). Thus, \(2\sigma=6\), so \(\sigma=3\).

Answer

a) \(\sigma=12\) b) \(\sigma=3\)

52528212

A normal random variable \(Y\) has mean \(\mu=100\). Use the \(68\text{-}95\text{-}99.7\) rule to find the standard deviation \(\sigma\) in each independent case. a) The probability that \(Y\) differs from its mean by more than \(15\) is approximately \(0.003\). b) \(P(Y\ge110)\approx0.159\)

Hints

- Translate a maximum deviation from the mean into a central interval. - Use symmetry to interpret a one-sided probability. - Match \(99.7\%\) with the correct number of standard deviations. - Relate tail probabilities to the central empirical-rule intervals.

Solution

1. a) The complement is \(P(|Y-100|\le15)\approx0.997\), which corresponds to the central three-standard-deviation interval. Thus, \(3\sigma=15\), so \(\sigma=5\). 2. b) The probability above one standard deviation over the mean is approximately \(0.5-0.683\div2=0.1585\). Therefore, \(110=100+\sigma\), so \(\sigma=10\).

Answer

a) \(\sigma=5\) b) \(\sigma=10\)

52529012

A normal random variable \(X\) has mean \(\mu=400\). Find \(\sigma\) in each independent case. a) \(P(|X-400|<32.8)\approx0.90\) b) \(P(X\le360.8)\approx0.025\)

Hints

- Recall the critical values for central \(90\%\) and \(95\%\) intervals. - Interpret \(|X-\mu|<d\) as a symmetric interval. - A left-tail probability of \(0.025\) leaves a central \(95\%\) interval between the two equal tails. - Match each probability with its standard score.

Solution

1. a) A central \(90\%\) interval has endpoints approximately \(1.64\) standard deviations from the mean. Therefore, \(1.64\sigma=32.8\), so \(\sigma=20\). 2. b) A left-tail probability of \(0.025\) corresponds to a standard score of approximately \(-1.96\). Thus, \(400-1.96\sigma=360.8\). 3. Solving gives \(1.96\sigma=39.2\), so \(\sigma=20\).

Answer

a) \(\sigma=20\) b) \(\sigma=20\)

52529512

A juice company labels its bottles as containing \(750\,\text{mL}\). The actual fill amount \(X\) is normally distributed with mean \(\mu=750\,\text{mL}\). For quality control, at least \(98\%\) of the bottles must contain between \(745\,\text{mL}\) and \(755\,\text{mL}\). Find the greatest allowable value of the standard deviation \(\sigma\). Round to four decimal places.

Hints

- Express the acceptable range as a symmetric interval around the mean. - Use the symmetry of the normal distribution to write the interval probability in terms of \(\Phi\). - Determine the standard-normal value that leaves \(1\%\) in each tail. - Solve the resulting inequality for \(\sigma\).

Solution

1. The requirement is \(P(745\le X\le755)\ge0.98\). 2. Standardizing and using symmetry gives \(2\Phi(5/\sigma)-1\ge0.98\). 3. Therefore, \(\Phi(5/\sigma)\ge0.99\). The \(0.99\) standard-normal quantile is approximately \(2.3263\). 4. Thus, \(5/\sigma\ge2.3263\), so \(\sigma\le5/2.3263\approx2.1493\,\text{mL}\).

Answer

The greatest allowable standard deviation is approximately \(2.1493\,\text{mL}\).

52529612

A factory produces precision parts whose length \(L\) is normally distributed with mean \(\mu=120\,\text{mm}\). A part is acceptable when its length is in the interval \([119.7\,\text{mm},120.3\,\text{mm}]\). Find the greatest allowable standard deviation \(\sigma\) if no more than \(1\%\) of the parts may be unacceptable. Round to four decimal places.

Hints

- Convert the maximum reject rate into the minimum acceptable-part probability. - Write the tolerance interval as a symmetric interval around the mean. - Standardize the interval and use the symmetry of the normal distribution. - Find the standard-normal quantile associated with a cumulative probability of \(0.995\).

Solution

1. If at most \(1\%\) are unacceptable, then at least \(99\%\) must be acceptable: \(P(119.7\le L\le120.3)\ge0.99\). 2. The tolerance interval is centered at the mean with radius \(0.3\,\text{mm}\). Standardizing gives \(P(-0.3/\sigma\le Z\le0.3/\sigma)\ge0.99\). 3. By symmetry, \(2\Phi(0.3/\sigma)-1\ge0.99\), so \(\Phi(0.3/\sigma)\ge0.995\). 4. Since \(z_{0.995}\approx2.5758\), we need \(0.3/\sigma\ge2.5758\). 5. Therefore, \(\sigma\le0.3/2.5758\approx0.1165\,\text{mm}\).

Answer

The greatest allowable standard deviation is approximately \(0.1165\,\text{mm}\).

52530712

For a normal random variable \(X\), \(P(X<60)=P(X>100)\approx0.0228\). Find the mean \(\mu\) and standard deviation \(\sigma\).

Hints

- Use symmetry to locate the mean between the two equal-tail boundaries. - Match the given tail probability with a familiar standard score. - Relate the tail boundaries to a central interval.

Solution

1. Equal tail probabilities occur at points equally far from the mean. Therefore, \(\mu=\frac{60+100}{2}=80\). 2. A one-sided tail probability of approximately \(0.0228\) corresponds to a standard score of about \(2\). Thus, the boundaries are approximately \(\mu-2\sigma\) and \(\mu+2\sigma\). 3. Using \(80-2\sigma=60\) gives \(2\sigma=20\), so \(\sigma=10\).

Answer

\(\mu=80\) and \(\sigma=10\)

52531212

The length \(X\) of an industrial pin is normally distributed with mean \(\mu=40.0\,\text{mm}\) and standard deviation \(\sigma=0.2\,\text{mm}\). a) Use the Empirical Rule to estimate \(P(39.8\le X\le40.2)\) and \(P(X<39.4)\). b) A normal distribution assigns some probability to negative values, but a physical length must be positive. Explain why the normal model is still appropriate in this situation. c) Explain how \(P(39.6\le X\le40.4)\) changes if the production process becomes more precise and the standard deviation decreases to \(0.1\,\text{mm}\), while the mean remains \(40.0\,\text{mm}\).

Hints

- Recall the Empirical Rule percentages within one, two, and three standard deviations. - Compare the distance from the mean to zero with the standard deviation. - Determine how many standard deviations each endpoint is from the mean before and after the change. - Consider how a smaller standard deviation changes the concentration of values around the mean.

Solution

1. a) The interval \([39.8, 40.2]\) is \([\mu-\sigma, \mu+\sigma]\). By the Empirical Rule, its probability is about \(68\%\). 2. The value \(39.4\) is \(\mu-3\sigma\). About \(0.3\%\) of values lie outside \(\mu\pm3\sigma\), so symmetry gives \(P(X<39.4)\approx0.15\%\). 3. b) Zero is \(\frac{40.0}{0.2}=200\) standard deviations below the mean. The normal model assigns an effectively zero probability to negative lengths, so this theoretical limitation has no practical effect. 4. c) With \(\sigma=0.2\), the interval \([39.6, 40.4]\) is \(\mu\pm2\sigma\), so its probability is about \(95\%\). 5. With \(\sigma=0.1\), the same interval is \(\mu\pm4\sigma\). The distribution is more concentrated around the mean, so the probability increases to nearly \(100\%\).

Answer

a) \(P(39.8\le X\le40.2)\approx68\%\); \(P(X<39.4)\approx0.15\%\) b) Zero is \(200\) standard deviations below the mean, so the modeled probability of a negative length is negligible. c) The probability increases from about \(95\%\) to nearly \(100\%\).

52531512

A normal random variable \(X\) has standard deviation \(\sigma=50\). In each independent case, use normal-distribution rules to find the mean \(\mu\). a) \(P(\mu\le X\le600)\approx0.477\) b) \(P(350\le X\le\mu)\approx0.3415\) c) \(P(X\ge750)\approx0.0015\)

Hints

- Match each probability with a one-, two-, or three-standard-deviation region. - Use symmetry to interpret areas that begin or end at the mean. - Determine how many standard deviations separate the boundary from the mean. - Identify and label each region conceptually before writing an equation.

Solution

1. a) The area from \(\mu\) to \(\mu+2\sigma\) is approximately \(0.477\). Thus, \(600=\mu+2\cdot50\), giving \(\mu=500\). 2. b) The area from \(\mu-\sigma\) to \(\mu\) is approximately \(0.3415\). Thus, \(350=\mu-50\), giving \(\mu=400\). 3. c) A right-tail probability of approximately \(0.0015\) begins at \(\mu+3\sigma\). Thus, \(750=\mu+3\cdot50\), giving \(\mu=600\).

Answer

a) \(\mu=500\) b) \(\mu=400\) c) \(\mu=600\)

52531612

A normal random variable \(X\) has standard deviation \(\sigma=20\). Find the mean \(\mu\) in each independent case using inverse-normal technology. Round each mean to the nearest tenth. a) \(P(X\le160)=0.1585\) b) \(P(X\le250)=0.9985\) c) \(P(X\ge170)=0.975\)

Hints

- Convert each stated probability into the corresponding standard-normal cumulative probability. - Use an inverse-normal value rather than treating rounded probabilities as exact Empirical-Rule landmarks. - Rearrange \(z=(x-\mu)/\sigma\) for \(\mu\). - Keep full precision until the final nearest-tenth rounding.

Solution

1. For part a), \(z=\Phi^{-1}(0.1585)\approx-1.00064\). Since \(\frac{160-\mu}{20}=z\), \(\mu=160-20z\approx180.0128\), so \(\mu\approx180.0\). 2. For part b), \(z=\Phi^{-1}(0.9985)\approx2.96774\). Since \(\frac{250-\mu}{20}=z\), \(\mu=250-20z\approx190.6452\), so \(\mu\approx190.6\). 3. For part c), \(P(X\le170)=0.025\), so \(z=\Phi^{-1}(0.025)\approx-1.95996\). Thus, \(\frac{170-\mu}{20}=z\), giving \(\mu\approx209.1993\), so \(\mu\approx209.2\).

Answer

a) \(\mu\approx180.0\) b) \(\mu\approx190.6\) c) \(\mu\approx209.2\)

52531712

A normal random variable \(X\) has probability density function \(\varphi(x)=\frac{1}{\sqrt{72\pi}}e^{-\frac{(x-40)^2}{72}}\). a) Use the \(68\text{-}95\text{-}99.7\) rule to estimate \(P(X\le34)\). b) Find the probability that \(X\) differs from its mean by no more than one standard deviation.

Hints

- Identify \(\mu\) and \(\sigma\) by comparing the density with the standard normal-density form. - Express \(34\) relative to the mean in standard deviation units. - Recall the percentages for central one-, two-, and three-standard-deviation intervals. - Use symmetry for the one-sided region in part a.

Solution

1. Compare the density with \(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\). This gives \(\mu=40\) and \(2\sigma^2=72\), so \(\sigma=6\). 2. a) Since \(34=\mu-\sigma\), the left-tail probability is approximately \((1-0.683)\div2=0.1585\). 3. b) The interval within one standard deviation of the mean is \([34, 46]\). By the empirical rule, its probability is approximately \(0.683\).

Answer

a) Approximately \(0.1585\), or \(15.85\%\) b) Approximately \(0.683\), or \(68.3\%\)

52531812

A normal random variable \(X\) has probability density function \(\varphi(x)=\frac{1}{\sqrt{50\pi}}e^{-\frac{(x+5)^2}{50}}\). a) Use the \(68\text{-}95\text{-}99.7\) rule to estimate \(P(X>10)\). b) Find the probability that \(X\) differs from its mean by at least two standard deviations.

Hints

- Find \(\mu\) and \(\sigma\) by matching the density to the standard form. - Express \(10\) as a number of standard deviations from the mean. - Interpret “at least two standard deviations” as a region outside a central interval. - Use symmetry of the normal distribution.

Solution

1. Comparing the density with \(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\) gives \(\mu=-5\) and \(2\sigma^2=50\), so \(\sigma=5\). 2. a) Since \(10=\mu+3\sigma\), the right-tail probability is approximately \((1-0.997)\div2=0.0015\). 3. b) The requested probability is outside the central two-standard-deviation interval. Thus, \(P(|X-\mu|\ge2\sigma)\approx1-0.954=0.046\).

Answer

a) Approximately \(0.0015\), or \(0.15\%\) b) Approximately \(0.046\), or \(4.6\%\)

52531912

A quality-control analyst wants a symmetric interval about the mean that contains about \(70\%\) of the measurements. a) For a standard normal random variable \(Z\), use a standard normal table or an inverse normal function to find \(k\), rounded to two decimal places, such that \(P(-k\le Z\le k)\approx0.70\). b) The weight \(G\) of flour packages is normally distributed with mean \(\mu=1000\,\text{g}\) and standard deviation \(\sigma=6\,\text{g}\). Using the value of \(k\) from part a), find an interval symmetric about \(\mu\) that contains about \(70\%\) of the package weights.

Hints

- Express the probability of a symmetric standard normal interval using \(\Phi\). - Determine the cumulative probability to the left of the upper endpoint. - Convert the standard normal endpoints back to the original scale.

Solution

1. a) For a symmetric standard normal interval, \(P(-k\le Z\le k)=2\Phi(k)-1\). Setting this equal to \(0.70\) gives \(\Phi(k)=0.85\). 2. A table or inverse normal function gives \(k\approx1.0364\), so \(k\approx1.04\) to two decimal places. 3. b) The corresponding interval for \(G\) is \([\mu-k\sigma, \mu+k\sigma]\). 4. Using the rounded value from part a), \(1000\pm1.04\cdot6=1000\pm6.24\). 5. The interval is \([993.76\,\text{g}, 1006.24\,\text{g}]\).

Answer

a) \(k\approx1.04\) b) \([993.76\,\text{g}, 1006.24\,\text{g}]\)

52532112

The fill amount \(X\), in milliliters, of a juice bottle is modeled by the normal density \(f(x)=\frac{1}{\sqrt{72\pi}}e^{-\frac{(x-500)^2}{72}}.\) 1. Find the mean \(\mu\) and standard deviation \(\sigma\). 2. Find the probability that a randomly selected bottle has a fill amount within one standard deviation of the mean. Round to four decimal places. 3. Find \(k\) such that \(P(X\le k)=0.90\). Round \(k\) to the nearest hundredth of a milliliter.

Hints

- Compare the given density with the standard normal-density form. - Identify which parts of the formula determine the mean and variance. - Standardize the central interval in part 2. - Use an inverse normal calculation for part 3.

Solution

1. Compare the density with \(\frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}.\) The center is \(\mu=500\), and \(2\sigma^2=72\), so \(\sigma=6.\) 2. The required interval is \([\mu-\sigma,\mu+\sigma]=[494{,}506].\) Therefore, \(P(494\le X\le506) =\Phi(1)-\Phi(-1) \approx0.6827.\) 3. Require \(\Phi\left(\frac{k-500}{6}\right)=0.90.\) The \(0.90\) standard-normal quantile is approximately \(1.2816\), so \(k=500+6(1.2816)\approx507.69\,\text{mL}.\)

Answer

1. \(\mu=500\,\text{mL}\), \(\sigma=6\,\text{mL}\). 2. \(0.6827\). 3. \(k\approx507.69\,\text{mL}\).

52532212

The mass \(X\), in grams, of a packaged food item is normally distributed as \(X\sim N(250,2^2).\) 1. A package is underweight if it weighs less than \(247\,\text{g}\). Find the probability that a randomly selected package is underweight. Round to four decimal places. 2. Find the narrowest interval centered at the mean, with integer endpoints in grams, that contains at least \(95\%\) of package masses.

Hints

- Standardize the underweight cutoff. - Express a centered interval as \([\mu-c,\mu+c]\). - Find the continuous radius required for \(95\%\), then impose the integer-endpoint constraint.

Solution

1. Standardizing gives \(z=\frac{247-250}{2}=-1.5.\) Therefore, \(P(X<247)=\Phi(-1.5)\approx0.0668.\) 2. Let the centered interval be \([250-c,250+c].\) We need \(2\Phi\left(\frac{c}{2}\right)-1\ge0.95.\) Thus, \(\frac{c}{2}\ge z_{0.975}\approx1.96,\) so \(c\ge3.92.\) Because the endpoints must be integers, the smallest allowable integer radius is \(c=4\). Therefore, the narrowest qualifying interval is \([246\,\text{g},254\,\text{g}].\)

Answer

1. \(0.0668\). 2. \([246\,\text{g},254\,\text{g}]\).

52533912

A normal random variable \(X\) has mean \(\mu=150\), and \(P(130\le X\le170)\approx0.9544.\) Use inverse-normal reasoning to estimate the standard deviation \(\sigma\). Round to two decimal places.

Hints

- Use the symmetry of the interval about the mean. - Convert the central probability into a one-sided cumulative probability. - Use an inverse-normal value rather than treating the rounded probability as an exact Empirical-Rule landmark.

Solution

1. The interval is symmetric about the mean with radius \(20\), so for \(z=\frac{20}{\sigma},\) we have \(P(-z\le Z\le z)\approx0.9544.\) 2. By symmetry, \(\Phi(z)\approx\frac{1+0.9544}{2}=0.9772.\) 3. Thus, \(z\approx\Phi^{-1}(0.9772)\approx1.9991.\) 4. Therefore, \(\sigma\approx\frac{20}{1.9991}\approx10.00.\)

Answer

\(\sigma\approx10.00\).

52534012

A normal random variable \(X\) has standard deviation \(\sigma=80\), and \(P(X>500)\approx0.95\). Find the mean \(\mu\), rounded to the nearest tenth.

Hints

- Decide whether the mean must be above or below \(500\). - Rewrite the right-tail probability as a left-tail probability. - Find the standard score associated with a \(5\%\) left tail.

Solution

1. The condition is equivalent to \(P(X\le500)\approx0.05\). 2. A left-tail probability of \(0.05\) corresponds to \(z\approx-1.6449\). 3. Standardizing gives \(-1.6449\approx(500-\mu)/80\). 4. Therefore, \(\mu\approx500-80(-1.6449)\approx631.6\).

Answer

\(\mu\approx631.6\).

52535112

A normal random variable \(X\) has mean \(\mu=120\) and standard deviation \(\sigma=15\). Find an interval \(I\) symmetric about \(\mu\) that contains each stated probability. Use \(z^*=1.00\) for \(68.3\%\), \(z^*=1.96\) for \(95.0\%\), and \(z^*=2.58\) for \(99.0\%\). Round noninteger endpoints to the nearest tenth. a) \(P(X\in I)\approx0.683\) b) \(P(X\in I)\approx0.950\) c) \(P(X\in I)\approx0.990\)

Hints

- Match each central probability with its supplied standard-normal critical value. - Use \([\mu-z^*\sigma,\mu+z^*\sigma]\). - Keep the interval centered at \(\mu\).

Solution

1. A symmetric central interval has the form \([\mu-z^*\sigma,\mu+z^*\sigma].\) 2. For \(68.3\%\), \(z^*=1.00\): \(I=[120-15{,}120+15]=[105{,}135].\) 3. For \(95.0\%\), \(z^*=1.96\): \(I=[120-1.96(15),120+1.96(15)] =[90.6{,}149.4].\) 4. For \(99.0\%\), \(z^*=2.58\): \(I=[120-2.58(15),120+2.58(15)] =[81.3{,}158.7].\)

Answer

a) \(I=[105{,}135]\) b) \(I=[90.6{,}149.4]\) c) \(I=[81.3{,}158.7]\)

52535212

The length \(X\) of metal pins produced by a factory is normally distributed with mean \(\mu=40\,\text{mm}\) and standard deviation \(\sigma=0.02\,\text{mm}\). Estimate each probability to three decimal places using normal-distribution rules. a) \(X\) lies in \(I_1=[39.9672\,\text{mm},40.0328\,\text{mm}]\). b) \(X\) lies in \(I_2=[39.96\,\text{mm},40.04\,\text{mm}]\). c) The length differs from the mean by more than \(0.06\,\text{mm}\).

Hints

- Find each interval’s radius and divide by \(\sigma\). - Determine how many standard deviations each boundary is from the mean. - In part c, use the complement of a central interval.

Solution

1. In part a, the interval radius is \(0.0328\,\text{mm}\), which is \(0.0328/0.02=1.64\) standard deviations. Thus, \(P(I_1)=P(-1.64\le Z\le1.64)\approx0.899\). 2. In part b, the interval radius is \(0.04\,\text{mm}=2\sigma\), so \(P(I_2)\approx0.954\). 3. In part c, \(0.06\,\text{mm}=3\sigma\), so the probability outside the central three-standard-deviation interval is approximately \(1-0.997=0.003\).

Answer

a) \(0.899\). b) \(0.954\). c) \(0.003\).

52535912

A random variable \(X\) is normally distributed with \(\mu=80\) and \(\sigma=5\). a) Find \(P(72\le X\le88)\), rounded to four decimal places. b) The standard deviation increases to \(8\), while the mean remains \(80\). Without recalculating, determine whether the probability over \([72,88]\) increases or decreases. Explain using the density curve. c) The mean changes to \(85\), while the standard deviation returns to \(5\). Explain how the probability over \([72,88]\) compares with the result from part a).

Hints

- Standardize the interval for part a). - A larger standard deviation makes a normal curve wider and lower. - Compare the fixed interval midpoint with the mean before and after the shift.

Solution

1. The standard scores are \(-1.6\) and \(1.6\). Thus, \(P(72\le X\le88)=2\Phi(1.6)-1\approx0.8904.\) 2. Increasing the standard deviation makes the curve wider and lower. Because the fixed interval remains centered at the mean, less probability lies inside it, so the probability decreases. 3. The midpoint of \([72,88]\) is \(80\). Moving the mean to \(85\) shifts the highest-density portion of the curve away from the center of the interval, so the probability also decreases relative to part a).

Answer

a) \(0.8904\). b) The probability decreases. c) The probability decreases.

52536612

A normal random variable \(Y\) has mean \(\mu=250\). Find \(\sigma\) in each independent case using inverse-normal values. Round to the nearest tenth. a) \(P(220\le Y\le280)=0.683\). b) \(P(Y\le274.6)=0.950\).

Hints

- Convert a symmetric central probability into an equal-tail cumulative probability. - Use an inverse-normal value rather than assuming an approximate probability equals an Empirical-Rule landmark exactly. - Write each boundary as \(\mu+z\sigma\) and solve for \(\sigma\).

Solution

1. In part a, symmetry gives \(P(-z\le Z\le z)=0.683\), so \(\Phi(z)=0.8415\) and \(z\approx1.00064\). 2. The interval radius is \(30\), so \(30/\sigma\approx1.00064\). Hence, \(\sigma\approx29.98\), which rounds to \(30.0\). 3. In part b, \(z_{0.950}\approx1.64485\). 4. Therefore, \((274.6-250)/\sigma=1.64485\), so \(\sigma\approx14.96\), which rounds to \(15.0\).

Answer

a) \(\sigma\approx30.0\). b) \(\sigma\approx15.0\).

52537712

A machine fills flour bags. The fill weight \(X\), in grams, is normally distributed with mean \(\mu=1006\) and standard deviation \(\sigma=4\). a) Find the probability that a randomly selected bag contains less than the labeled weight of \(1000\,\text{g}\). Round to four decimal places. b) The company wants exactly \(0.5\%\) of the bags to contain less than \(1000\,\text{g}\). Find the required standard deviation if the mean remains \(1006\,\text{g}\). Round to the nearest hundredth. c) Find the required mean if the standard deviation remains \(4\,\text{g}\). Round to the nearest tenth.

Hints

- Standardize the labeled weight in part a). - Find the standard-normal quantile for a \(0.5\%\) left tail. - Rearrange the standard-score equation for the unknown parameter.

Solution

1. For part a), \(z=\frac{1000-1006}{4}=-1.5,\) so \(P(X<1000)=\Phi(-1.5)\approx0.0668.\) 2. For a lower-tail probability of \(0.005\), the standard-normal quantile is approximately \(z_{0.005}=-2.5758.\) With \(\mu=1006\), \(\frac{1000-1006}{\sigma}=-2.5758,\) so \(\sigma\approx2.33\,\text{g}.\) 3. With \(\sigma=4\), \(\frac{1000-\mu}{4}=-2.5758,\) giving \(\mu\approx1010.3\,\text{g}.\)

Answer

a) \(0.0668\). b) \(\sigma\approx2.33\,\text{g}\). c) \(\mu\approx1010.3\,\text{g}\).

52538212

The fill weight \(X\) of a brand of coffee pods is approximately normally distributed with mean \(\mu=6.2\,\text{g}\) and standard deviation \(\sigma=0.15\,\text{g}\). Find each probability, rounded to four decimal places. 1) A randomly selected pod has a fill weight between \(6.0\,\text{g}\) and \(6.4\,\text{g}\). 2) A pod’s fill weight differs from the mean by more than \(0.3\,\text{g}\).

Hints

- Compare each endpoint with the mean and check whether the interval is symmetric. - Translate “differs by more than” into two tails of the distribution. - Express each deviation in standard-deviation units.

Solution

1. For \(6.0\le X\le6.4\), the standard scores are \(-4/3\) and \(4/3\). Thus, \(P(6.0\le X\le6.4)=\Phi(4/3)-\Phi(-4/3)\approx0.8176\). 2. A deviation greater than \(0.3\,\text{g}\) means \(X<5.9\) or \(X>6.5\). Since \(0.3/0.15=2\), this is the area outside \([-2,2]\). 3. Therefore, \(P(|X-6.2|>0.3)=1-[\Phi(2)-\Phi(-2)]\approx0.0455\).

Answer

1) \(0.8176\). 2) \(0.0455\).

52538312

A normal random variable \(X\) has mean \(\mu=40\) and standard deviation \(\sigma=4\). Find the positive value \(c\) in each independent case using inverse-normal technology. Round to the nearest hundredth. a) \(P(X\le c)=0.1585\). b) \(P(32\le X\le c)=0.954\). c) \(P(X\ge c)=0.005\).

Hints

- Translate each probability statement into a cumulative normal probability. - In part b, account for the nonzero area below the lower endpoint rather than assuming the interval is exactly central. - Use inverse-normal values and keep full precision until the final rounding.

Solution

1. For part a, \(z=\Phi^{-1}(0.1585)\approx-1.00064\). Thus, \(c=40+4z\approx35.9974\), so \(c\approx36.00\). 2. For part b, \(32\) has standard score \(-2\). The equation is \(\Phi((c-40)/4)-\Phi(-2)=0.954\). 3. Hence, \(\Phi((c-40)/4)=0.954+\Phi(-2)\), which gives \((c-40)/4\approx1.99083\). Therefore, \(c\approx47.96\). 4. For part c, \(P(X\le c)=0.995\), so \(z=\Phi^{-1}(0.995)\approx2.57583\). 5. Therefore, \(c=40+4(2.57583)\approx50.30\).

Answer

a) \(c\approx36.00\). b) \(c\approx47.96\). c) \(c\approx50.30\).

52574212

Studies show that a particular side effect occurs in \(5\%\) of patients. Assume patient outcomes are independent and the side-effect probability is constant. Let \(X\) be the number of patients who experience the side effect in a group of \(n\) patients. a) Find the sample size \(n\) for which the mean number of patients with the side effect is exactly \(20\). b) For this sample size, find the standard deviation \(\sigma\). c) Check whether a normal approximation is appropriate using the success-failure condition \(np\ge10\) and \(n(1-p)\ge10\).

Hints

- Use the binomial mean formula to find the sample size. - Then use the binomial standard-deviation formula. - Substitute the values into both parts of the success-failure condition.

Solution

1. Since \(E(X)=np=20\) and \(p=0.05\), \(n=\frac{20}{0.05}=400\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{400\cdot0.05\cdot0.95}=\sqrt{19}\approx4.36\). 3. For the success-failure condition, \(np=400\cdot0.05=20\) and \(n(1-p)=400\cdot0.95=380\). 4. Both values are at least \(10\), so a normal approximation is appropriate.

Answer

a) \(n=400\) b) \(\sigma=\sqrt{19}\approx4.36\) c) Yes. Since \(np=20\) and \(n(1-p)=380\), both parts of the success-failure condition are satisfied.

52694212

The fill amount \(X\), in milliliters, of a manufacturer's juice bottles is normally distributed with mean \(\mu=1000\) and standard deviation \(\sigma=8\). a) Use the Empirical Rule to estimate the percentage of bottles containing less than \(984\,\text{mL}\). b) Find the percentage of bottles whose fill amounts differ from the mean by more than \(12\,\text{mL}\). Round to two decimal places. c) The manufacturer wants to set a minimum fill amount \(m\) that only \(1\%\) of the bottles fall below. Find \(m\) to the nearest hundredth of a milliliter.

Hints

- Express the cutoff in part a) as a number of standard deviations from the mean. - Include both tails when the deviation may be above or below the mean. - Use an inverse normal value for the first percentile.

Solution

1. Since \(984=1000-2(8)=\mu-2\sigma,\) the Empirical Rule places about \(5\%\) outside the central two-standard-deviation interval. By symmetry, about \(2.5\%\) is in the lower tail. 2. A deviation greater than \(12\,\text{mL}\) corresponds to \(|Z|>\frac{12}{8}=1.5.\) Therefore, \(P(|X-1000|>12)=2[1-\Phi(1.5)]\approx0.1336,\) or \(13.36\%\). 3. For part c), require \(P(X<m)=0.01.\) The first-percentile standard-normal value is approximately \(z_{0.01}=-2.3263.\) Hence, \(m=1000+8(-2.3263)\approx981.39\,\text{mL}.\)

Answer

a) Approximately \(2.5\%\). b) \(13.36\%\). c) \(m\approx981.39\,\text{mL}\).

53112112

A pharmaceutical company produces tablets. Historically, \(15\%\) of the tablets have an active-ingredient level slightly above the target. A batch of \(600\) tablets is inspected. Use a normal approximation to the binomial distribution to find the radius \(r\) of a symmetric interval about the expected number of above-target tablets that contains approximately \(95\%\) of the count distribution. Round \(r\) to two decimal places.

Hints

- Find the binomial mean and standard deviation first. - Check the large-count condition before using the normal approximation. - Multiply the standard deviation by the central-\(95\%\) critical value.

Solution

1. Let \(X\sim\operatorname{Bin}(600,0.15).\) The mean is \(\mu=np=90.\) 2. The standard deviation is \(\sigma=\sqrt{np(1-p)} =\sqrt{600(0.15)(0.85)} =\sqrt{76.5} \approx8.746.\) The large-count condition is satisfied. 3. A central \(95\%\) normal interval uses \(z^*\approx1.96.\) Therefore, \(r=z^*\sigma\approx1.96(8.746)\approx17.14.\)

Answer

\(r\approx17.14\), giving the approximate continuous interval \([90-17.14,90+17.14]=[72.86{,}107.14].\)

53112712

A spinner has \(10\) equal sections, \(3\) of which are red. The spinner is spun independently \(400\) times, and \(X\) is the number of red results. 1. Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). Give \(\sigma\) exactly and to two decimal places. 2. Find the endpoints of \([\mu-2\sigma,\mu+2\sigma]\) to two decimal places and list the possible integer values of \(X\) in the interval. 3. Using a normal approximation and the Empirical Rule, estimate the probability that \(X\) lies in this interval.

Hints

- Use the binomial mean and standard-deviation formulas. - Remember that the binomial random variable takes integer values. - Check the large-count condition before invoking a normal approximation. - Match a two-standard-deviation interval with the Empirical Rule.

Solution

1. Since \(X\sim\operatorname{Bin}(400,0.3),\) \(\mu=np=120\) and \(\sigma=\sqrt{np(1-p)}=\sqrt{84}\approx9.17.\) 2. The radius is \(2\sigma=2\sqrt{84}\approx18.33,\) so the continuous interval is approximately \([101.67{,}138.33].\) The attainable integer values are \(102{,}103,\ldots,138\). 3. The large-count condition is satisfied, and the Empirical Rule gives approximately \(95.4\%\) within two standard deviations for the normal approximation.

Answer

1. \(\mu=120\), \(\sigma=\sqrt{84}\approx9.17\). 2. Approximately \([101.67{,}138.33]\); integer values \(102\) through \(138\). 3. Approximately \(95.4\%\).

53112812

A binomial random variable \(X\) has \(n=100\) and \(p=0.4\). Determine whether the probability of the one-standard-deviation interval about the mean is closer to \(68\%\) or \(70\%\). First find the integer values in the interval, then use a normal approximation with continuity correction. Report the approximate probability to the nearest tenth of a percent.

Hints

- Find the continuous one-standard-deviation interval and identify its integer values. - Extend the integer interval by \(0.5\) at each end for the continuity correction. - Compare the final percentage with both proposed benchmarks.

Solution

1. The mean is \(\mu=np=40,\) and the standard deviation is \(\sigma=\sqrt{100(0.4)(0.6)}=\sqrt{24}\approx4.899.\) 2. The one-standard-deviation interval is approximately \([35.101,44.899],\) so its integer values are \(36\) through \(44\). 3. With continuity correction, approximate \(P(36\le X\le44)\) by the normal probability over \(35.5\le Y\le44.5,\) where \(Y\) has mean \(40\) and standard deviation \(\sqrt{24}\). 4. The standard scores are approximately \(-0.919\) and \(0.919\), giving \(P(36\le X\le44)\approx0.6417=64.2\%.\) 5. This is \(3.8\) percentage points from \(68\%\) and \(5.8\) percentage points from \(70\%\), so it is closer to \(68\%\).

Answer

The integer values are \(36\) through \(44\). The normal approximation gives about \(64.2\%\), which is closer to \(68\%\).

53113112

A pharmaceutical company reports that \(15\%\) of patients experience side effects from a certain tablet. In a study of \(400\) patients, let \(X\) be the number who experience side effects. 1. Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\), and check whether a normal approximation is appropriate. 2. Use the binomial distribution to find the probability that \(X\) lies within \(1.96\sigma\) of its mean. Round to four decimal places. 3. Compare your result with the corresponding central normal-distribution value.

Hints

- Use the binomial formulas for mean and standard deviation. - Check both expected successes and expected failures. - Identify the integer values inside the continuous \(1.96\sigma\) interval. - Compare the exact binomial probability with the normal benchmark.

Solution

1. Since \(X\sim\operatorname{Bin}(400,0.15),\) \(\mu=60,\qquad \sigma=\sqrt{51}\approx7.14.\) The large-count condition is satisfied because \(np=60\) and \(n(1-p)=340\). 2. The radius is \(1.96\sigma\approx13.997,\) so the integer values inside the interval are \(47\) through \(73\). 3. Using the binomial distribution, \(P(47\le X\le73)\approx0.9417.\) 4. The exact binomial probability, about \(94.17\%\), is close to the central normal value of \(95\%\), but not identical because the binomial distribution is discrete.

Answer

1. \(\mu=60\), \(\sigma=\sqrt{51}\approx7.14\); the normal approximation is appropriate. 2. \(P(47\le X\le73)\approx0.9417\). 3. \(94.17\%\) is close to \(95\%\).

53113912

For each binomial random variable \(X\), use a normal approximation to find the requested symmetric interval about the binomial mean \(\mu\). Give continuous endpoints to two decimal places and then list the integer outcomes inside the interval. a) \(n=600\), \(p=0.20\): central \(90\%\) interval. b) \(n=400\), \(p=0.70\): central \(95\%\) interval.

Hints

- Find the binomial mean and standard deviation. - Check the large-count condition before using a normal approximation. - Use the appropriate central-normal critical value. - Distinguish the continuous approximation interval from the attainable integer counts.

Solution

1. In part a, \(\mu=np=120\) and \(\sigma=\sqrt{np(1-p)}=\sqrt{96}\approx9.798\). The large-count condition is satisfied. 2. For a central \(90\%\) interval, \(z^*\approx1.645\). Thus, \(120\pm1.645(9.798)\approx[103.88,136.12]\), containing integer outcomes \(104\) through \(136\). 3. In part b, \(\mu=280\) and \(\sigma=\sqrt{84}\approx9.165\). The large-count condition is satisfied. 4. For a central \(95\%\) interval, \(z^*\approx1.96\). Thus, \(280\pm1.96(9.165)\approx[262.04,297.96]\), containing integer outcomes \(263\) through \(297\).

Answer

a) Continuous approximation \([103.88,136.12]\); integer outcomes \(104\) through \(136\). b) Continuous approximation \([262.04,297.96]\); integer outcomes \(263\) through \(297\).

53114012

A binomial random variable \(X\) has parameters \(n=1250\) and \(p=0.16\). a) Use a normal approximation to find a central \(95\%\) interval about the binomial mean. Give continuous endpoints to two decimal places and list the integer outcomes inside the interval. b) Without further calculation, decide whether a central \(99\%\) interval about the same mean would be wider or narrower than the \(95\%\) interval. Explain.

Hints

- Find the binomial mean and standard deviation. - Check the large-count condition before using a normal approximation. - Use the critical value for a central \(95\%\) normal interval. - Relate greater captured probability to a larger critical value.

Solution

1. The binomial mean is \(\mu=np=200\), and the standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{168}\approx12.961\). The large-count condition is satisfied. 2. For a central \(95\%\) interval, \(z^*\approx1.96\). 3. Therefore, \(200\pm1.96(12.961)\approx[174.60,225.40]\), containing integer outcomes \(175\) through \(225\). 4. A central \(99\%\) interval must include more probability, so it uses a larger critical value and is wider.

Answer

a) Continuous approximation \([174.60,225.40]\); integer outcomes \(175\) through \(225\). b) The \(99\%\) interval is wider.

53114412

A manufacturer produces screws, and each screw independently has probability \(0.04\) of being defective. A quality-control sample contains \(1000\) screws. Use a normal approximation to find a symmetric interval about the binomial mean that contains the number of defective screws with probability approximately \(0.95\). Give the continuous endpoints to two decimal places and the possible integer counts in the interval.

Hints

- Find the binomial mean and standard deviation. - Check the large-count condition before using the normal approximation. - Use the critical value for a central \(95\%\) normal interval. - Distinguish the continuous approximation from the attainable integer counts.

Solution

1. Let \(X\sim\operatorname{Bin}(1000,0.04)\). The mean is \(\mu=np=40\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{38.4}\approx6.197\). Since \(np=40\) and \(n(1-p)=960\), the normal approximation is appropriate. 3. A central \(95\%\) interval uses \(z^*\approx1.96\). 4. The radius is \(1.96(6.197\ldots)\approx12.15\), so the continuous interval is approximately \([27.85,52.15]\). 5. The integer counts inside this interval are \(28\) through \(52\).

Answer

Continuous approximation: \([27.85,52.15]\). Integer counts: \(28\) through \(52\).

53119412

At a distribution center, each package independently has probability \(0.05\) of requiring manual processing after the first scan. On one day, \(2400\) packages are processed. a) How many packages are expected to require manual processing? b) Find the three-standard-deviation interval about the binomial mean. Give the continuous endpoints to two decimal places, list the integer counts in the interval, and state the approximate probability covered by the interval.

Hints

- Interpret the binomial mean as an expected count. - Use the binomial standard-deviation formula. - Build an interval symmetric about the mean. - Check the large-count condition before applying the Empirical Rule to the normal approximation.

Solution

1. Let \(X\sim\operatorname{Bin}(2400,0.05)\). The mean is \(\mu=np=120\). 2. The standard deviation is \(\sigma=\sqrt{2400(0.05)(0.95)}=\sqrt{114}\approx10.68\). 3. The radius is \(3\sigma\approx32.03\), so the continuous interval is approximately \([87.97,152.03]\), containing integer counts \(88\) through \(152\). 4. The large-count condition is satisfied, so a normal approximation is appropriate. 5. By the Empirical Rule, a three-standard-deviation interval contains approximately \(99.7\%\) of the normal approximation.

Answer

a) \(120\) packages. b) Continuous interval approximately \([87.97,152.03]\); integer counts \(88\) through \(152\); approximate coverage \(99.7\%\).

53119912

A pharmaceutical company produces tablet packages, and each package independently has probability \(0.015\) of missing the target weight. A quality-control sample contains \(6000\) packages. a) Find the mean \(\mu\) and standard deviation \(\sigma\) of the number of packages that miss the target weight. Give \(\sigma\) to two decimal places. b) Use a normal approximation to find central \(95\%\) and \(99\%\) intervals about the binomial mean. Give continuous endpoints to two decimal places and the possible integer counts.

Hints

- Identify the binomial parameters. - Use the formulas for the binomial mean and standard deviation. - Check the large-count condition before using a normal approximation. - Match central \(95\%\) and \(99\%\) intervals with their normal critical values.

Solution

1. Let \(X\sim\operatorname{Bin}(6000,0.015)\). The mean is \(\mu=np=90\). 2. The standard deviation is \(\sigma=\sqrt{6000(0.015)(0.985)}=\sqrt{88.65}\approx9.42\). The large-count condition is satisfied. 3. For a central \(95\%\) interval, use \(z^*\approx1.96\): \(90\pm1.96(9.415\ldots)\approx[71.55,108.45]\), containing integer counts \(72\) through \(108\). 4. For a central \(99\%\) interval, use \(z^*\approx2.576\): \(90\pm2.576(9.415\ldots)\approx[65.75,114.25]\), containing integer counts \(66\) through \(114\).

Answer

a) \(\mu=90\), \(\sigma\approx9.42\). b) \(95\%\): continuous interval approximately \([71.55,108.45]\), integer counts \(72\) through \(108\). \(99\%\): continuous interval approximately \([65.75,114.25]\), integer counts \(66\) through \(114\).

53120012

In a large city, \(20\%\) of households subscribe to a particular streaming service. A random sample contains \(n=1600\) households. a) Find the two-standard-deviation interval for the number of subscribing households and state its approximate probability. b) In the sample, \(360\) households report having a subscription. Use the three-standard-deviation interval to decide whether this result is unusually far from the expected count.

Hints

- Interpret a two- or three-standard-deviation interval as a distance from the mean. - Recall the empirical-rule probabilities. - Compare the observed count with the calculated endpoints.

Solution

1. The mean is \(\mu=np=1600\cdot0.20=320\), and the standard deviation is \(\sigma=\sqrt{1600\cdot0.20\cdot0.80}=\sqrt{256}=16\). The large-count condition is satisfied because \(np=320\) and \(n(1-p)=1280\), so a normal approximation is appropriate. 2. a) The two-standard-deviation interval is \([320-2\cdot16, 320+2\cdot16]=[288, 352]\). By the empirical rule, its probability is approximately \(95.4\%\). 3. b) The three-standard-deviation interval is \([320-3\cdot16, 320+3\cdot16]=[272, 368]\). 4. Since \(360\) lies inside this interval, it is not unusually far from the expected count by the three-standard-deviation criterion.

Answer

a) \([288, 352]\); approximately \(95.4\%\) b) \(360\) is inside \([272, 368]\), so it is not unusual by this criterion.

53120112

A manufacturer knows that each precision component independently has probability \(0.04\) of having a minor cosmetic defect. A shipment contains \(2500\) components. Use a normal approximation to find a central \(95\%\) interval about the binomial mean for the number of defective components. Give continuous endpoints to two decimal places and the possible integer counts.

Hints

- Identify the binomial model. - Find its mean and standard deviation. - Check the large-count condition. - Use the critical value for a central \(95\%\) normal interval.

Solution

1. Let \(X\sim\operatorname{Bin}(2500,0.04)\). The mean is \(\mu=np=100\). 2. The standard deviation is \(\sigma=\sqrt{2500(0.04)(0.96)}=\sqrt{96}\approx9.798\). The large-count condition is satisfied. 3. A central \(95\%\) interval uses \(z^*\approx1.96\). Thus, \(100\pm1.96(9.798\ldots)\approx[80.80,119.20]\). 4. The integer counts inside the interval are \(81\) through \(119\).

Answer

Continuous approximation: \([80.80,119.20]\). Integer counts: \(81\) through \(119\).

53120212

In a large city, each randomly selected resident can be treated as independently using public transit regularly with probability \(0.60\). A sample contains \(600\) residents. Use a normal approximation to find central \(90\%\) and \(99\%\) intervals about the binomial mean for the number of regular public-transit users. Give continuous endpoints to two decimal places and the possible integer counts.

Hints

- Find the binomial mean and standard deviation. - Check the large-count condition. - Use the critical values for central \(90\%\) and \(99\%\) normal intervals. - Distinguish continuous endpoints from attainable integer counts.

Solution

1. Let \(X\sim\operatorname{Bin}(600,0.60)\). The mean is \(\mu=360\), and the standard deviation is \(\sigma=\sqrt{600(0.60)(0.40)}=12\). The large-count condition is satisfied. 2. For a central \(90\%\) interval, use \(z^*\approx1.645\): \(360\pm1.645(12)\approx[340.26,379.74]\), containing integer counts \(341\) through \(379\). 3. For a central \(99\%\) interval, use \(z^*\approx2.576\): \(360\pm2.576(12)\approx[329.09,390.91]\), containing integer counts \(330\) through \(390\).

Answer

Central \(90\%\): continuous interval approximately \([340.26,379.74]\), integer counts \(341\) through \(379\). Central \(99\%\): continuous interval approximately \([329.09,390.91]\), integer counts \(330\) through \(390\).

53120312

A radio station claims that each person in a large city independently listens regularly to its morning show with probability \(0.20\). A sample contains \(1600\) people. 1. Assuming the claim is correct, find the mean \(\mu\) and standard deviation \(\sigma\) of the number of regular listeners in the sample. 2. Use a normal approximation to find a symmetric central \(95\%\) interval about the binomial mean. Give continuous endpoints to two decimal places and the integer counts inside the interval. 3. Estimate the probability that the sample count differs from the mean by more than \(32\), using the Empirical Rule.

Hints

- Identify the binomial parameters first. - Check the large-count condition before using a normal approximation. - Distinguish a central \(95\%\) critical-value interval from a two-standard-deviation Empirical-Rule interval. - A deviation from the mean can occur in either direction.

Solution

1. Let \(X\sim\operatorname{Bin}(1600,0.20)\). Then \(\mu=np=320\) and \(\sigma=\sqrt{1600(0.20)(0.80)}=16\). The large-count condition is satisfied. 2. A central \(95\%\) interval uses \(z^*\approx1.96\), giving \(320\pm1.96(16)=[288.64,351.36]\). The integer counts inside are \(289\) through \(351\). 3. A deviation of \(32\) is \(2\sigma\). The Empirical Rule places approximately \(95.4\%\) within two standard deviations, so the probability of a larger deviation is approximately \(4.6\%\).

Answer

1. \(\mu=320\), \(\sigma=16\). 2. Continuous interval \([288.64,351.36]\); integer counts \(289\) through \(351\). 3. Approximately \(4.6\%\).

53120412

Each microchip produced by a manufacturer independently has probability \(0.10\) of being defective. A quality-control sample contains \(900\) chips. Let \(X\) be the number of defective chips. Complete each statement using the binomial mean and standard deviation together with normal-distribution rules. a) With probability approximately \(32\%\), the number of defective chips differs from its mean by more than \(\ldots\). b) In only about \(5\%\) of samples, the number of defective chips lies outside the range from \(\ldots\) to \(\ldots\). Give the integer counts inside the corresponding central \(95\%\) interval. c) The probability that the number of defective chips differs from its mean by more than \(27\) is approximately \(\ldots\).

Hints

- Find the binomial mean and standard deviation first. - Match approximately \(32\%\) outside with the complement of the central one-standard-deviation region. - For a central \(95\%\) approximation, use \(1.96\) standard deviations. - Compare the given deviation in part c with \(\sigma\).

Solution

1. \(X\sim\operatorname{Bin}(900,0.10)\), so \(\mu=np=90\) and \(\sigma=\sqrt{900(0.10)(0.90)}=9\). The large-count condition is satisfied. 2. The probability outside the central one-standard-deviation interval is about \(31.7\%\), so the deviation in part a is \(9\). 3. A central \(95\%\) interval uses \(z^*\approx1.96\): \(90\pm1.96(9)=[72.36,107.64]\). The integer counts inside are \(73\) through \(107\). 4. Since \(27=3\sigma\), the probability of a deviation greater than \(27\) is approximately the probability outside three standard deviations, or \(0.3\%\).

Answer

a) \(9\). b) \(73\) through \(107\). c) Approximately \(0.3\%\).

53120512

A fair six-sided die is rolled independently \(1800\) times. Let \(X\) be the number of sixes. For each continuous interval below, find the integer values of \(X\) that actually lie inside it. 1. \([\mu-\sigma,\mu+\sigma]\) 2. \([\mu-2\sigma,\mu+2\sigma]\) 3. \([\mu-3\sigma,\mu+3\sigma]\) Also state the corresponding approximate normal probabilities from the \(68\text{-}95\text{-}99.7\) rule.

Hints

- Find the binomial mean and standard deviation. - Compute each continuous interval first. - To list integer values that lie inside a continuous interval, use the first integer at or above the lower endpoint and the last integer at or below the upper endpoint. - Do not enlarge the interval by rounding outward.

Solution

1. \(X\sim\operatorname{Bin}(1800,1/6)\), so \(\mu=300\) and \(\sigma=\sqrt{250}\approx15.811\). The large-count condition is satisfied. 2. The one-standard-deviation interval is approximately \([284.189,315.811]\). The integer values inside it are \(285\) through \(315\), with approximate normal probability \(68.3\%\). 3. The two-standard-deviation interval is approximately \([268.377,331.623]\). The integer values inside it are \(269\) through \(331\), with approximate normal probability \(95.4\%\). 4. The three-standard-deviation interval is approximately \([252.566,347.434]\). The integer values inside it are \(253\) through \(347\), with approximate normal probability \(99.7\%\).

Answer

1. \(285\) through \(315\); approximately \(68.3\%\). 2. \(269\) through \(331\); approximately \(95.4\%\). 3. \(253\) through \(347\); approximately \(99.7\%\).

53120612

Each microchip produced by a factory independently has probability \(0.03\) of being defective. A quality-control sample contains \(2000\) chips, and \(X\) is the number of defective chips. a) Check whether a normal approximation to the binomial distribution is appropriate. b) Find the integer counts in the interval \([\mu-2\sigma,\mu+2\sigma]\), and state the corresponding approximate normal probability.

Hints

- Check the expected numbers of successes and failures. - Match a two-standard-deviation interval with the Empirical Rule. - List only integer counts that lie inside the continuous interval.

Solution

1. \(X\sim\operatorname{Bin}(2000,0.03)\), so \(\mu=np=60\) and \(\sigma=\sqrt{2000(0.03)(0.97)}=\sqrt{58.2}\approx7.629\). 2. The large-count condition is satisfied because \(np=60\) and \(n(1-p)=1940\), so a normal approximation is appropriate. 3. The continuous two-standard-deviation interval is approximately \(60\pm2(7.629)=[44.742,75.258]\). 4. The integer counts inside are \(45\) through \(75\). The Empirical Rule gives approximate normal coverage \(95.4\%\).

Answer

a) Yes; the large-count condition is satisfied. b) \(45\) through \(75\) defective chips; approximately \(95.4\%\).

53120812

A pharmaceutical company reports that each patient independently has probability \(0.025\) of experiencing a mild injection-site reaction after receiving a vaccine. A clinical study includes \(4000\) patients. a) Find the continuous three-standard-deviation interval about the binomial mean for the number of patients with the reaction, and list the integer counts inside it. b) In the study, \(160\) patients have the reaction. Using the stated three-standard-deviation criterion, assess whether this count is consistent with the company's reported rate.

Hints

- Find the binomial mean and standard deviation. - Check the large-count condition. - Build the interval with radius \(3\sigma\). - Apply the criterion stated in the question rather than treating “consistent” as an undefined judgment.

Solution

1. Let \(X\sim\operatorname{Bin}(4000,0.025)\). The mean is \(\mu=100\), and the standard deviation is \(\sigma=\sqrt{4000(0.025)(0.975)}=\sqrt{97.5}\approx9.874\). The large-count condition is satisfied. 2. The continuous three-standard-deviation interval is \(100\pm3(9.874\ldots)\approx[70.38,129.62]\). The integer counts inside are \(71\) through \(129\). 3. The observed count \(160\) is above this interval. 4. Under the stated criterion, an observation outside three standard deviations is treated as inconsistent with the reported rate; such outcomes have total normal-approximation probability about \(0.3\%\).

Answer

a) Continuous interval approximately \([70.38,129.62]\); integer counts \(71\) through \(129\). b) No. The count \(160\) lies outside the three-standard-deviation interval.

53121012

A pharmaceutical company claims that a new medication produces the desired response in \(85\%\) of patients. A random sample of \(600\) patients is selected. Let \(\hat p\) be the sample proportion who respond. a) Assuming the claim is correct, explain why the sampling distribution of \(\hat p\) is approximately normal. Find its mean and standard deviation. b) Use a normal approximation to find a central \(99\%\) interval for \(\hat p\). Round the endpoints to four decimal places. c) In the study, \(486\) patients respond. Determine whether the observed sample proportion is inside the interval from part b), and interpret the result relative to the claimed \(85\%\) response rate.

Hints

- Work directly with the sampling distribution of the sample proportion. - Check the large-count condition before using a normal approximation. - Use \(\sqrt{p(1-p)/n}\) for the standard deviation of \(\hat p\). - Compare the observed sample proportion with the predicted interval.

Solution

1. Under the claim, \(p=0.85,\qquad n=600.\) The large-count values are \(np=510,\qquad n(1-p)=90,\) so the sampling distribution of \(\hat p\) is approximately normal. 2. Its mean is \(\mu_{\hat p}=p=0.85,\) and its standard deviation is \(\sigma_{\hat p} =\sqrt{\frac{p(1-p)}{n}} =\sqrt{\frac{0.85(0.15)}{600}} \approx0.01458.\) 3. A central \(99\%\) interval uses \(z^*\approx2.576\): \(0.85\pm2.576(0.01458) \approx[0.8125,0.8875].\) 4. The observed sample proportion is \(\hat p=\frac{486}{600}=0.81.\) Since \(0.81<0.8125\), the observed proportion lies just below the central \(99\%\) interval under the claim.

Answer

a) Approximately normal because \(np=510\) and \(n(1-p)=90\). Mean \(0.85\); standard deviation approximately \(0.01458\). b) Approximately \([0.8125,0.8875]\). c) \(\hat p=0.81\), which lies below the interval; the observed response rate is unusually low relative to the \(85\%\) claim at this prediction level.

53274812

A spice company fills small pepper packets with a target weight of \(12\,\text{g}\). The actual fill weights from two machines are normally distributed. The graph shows their density curves. Filled markers show the peaks, and open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) for each machine from the graph. Explain how the marked points support your values. b) A packet is underfilled if it weighs less than \(12\,\text{g}\). Find the standardized score of \(12\,\text{g}\) for each machine, then find the underfilled proportion for each machine to four decimal places. Which machine produces the greater underfilled proportion? c) Which machine has fill weights more tightly concentrated around its own mean? Justify your answer from the graph and the standard deviations.

Figure for problem 532748

Hints

- A normal density reaches its maximum at the mean. - Its inflection points occur at \(\mu-\sigma\) and \(\mu+\sigma\). - Standardize the common \(12\,\text{g}\) cutoff separately for each machine. - A smaller standard deviation gives a narrower, taller normal curve.

Solution

1. The filled peak of \(f_A\) is at \(x=10\), and its open inflection markers are at \(x=8\) and \(x=12\). Hence, \(\mu_A=10,\qquad \sigma_A=2.\) 2. The filled peak of \(f_B\) is at \(x=14\), and its open inflection markers are at \(x=13\) and \(x=15\). Hence, \(\mu_B=14,\qquad \sigma_B=1.\) 3. For Machine A, \(z_A=\frac{12-10}{2}=1,\) so \(P_A(X<12)=\Phi(1)\approx0.8413.\) 4. For Machine B, \(z_B=\frac{12-14}{1}=-2,\) so \(P_B(X<12)=\Phi(-2)\approx0.0228.\) Therefore, Machine A has the greater underfilled proportion. 5. Machine B is more tightly concentrated around its mean because \(\sigma_B=1<2=\sigma_A\); its density curve is correspondingly narrower and taller.

Answer

a) Machine A: \(\mu_A=10\,\text{g}\), \(\sigma_A=2\,\text{g}\). Machine B: \(\mu_B=14\,\text{g}\), \(\sigma_B=1\,\text{g}\). b) \(z_A=1\), underfilled proportion \(0.8413\); \(z_B=-2\), underfilled proportion \(0.0228\). Machine A has the greater underfilled proportion. c) Machine B is more tightly concentrated because it has the smaller standard deviation.

53275112

The graph shows the probability density function of a normal random variable \(X\). The filled marker shows the peak, and the two open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. Briefly justify your answers. b) Use the \(68\text{-}95\text{-}99.7\) rule to estimate each probability. 1. \(P(3\le X\le5)\) 2. \(P(2\le X\le6)\) 3. \(P(X\ge5)\)

Figure for problem 532751

Hints

- Read the peak and marked inflection points from the graph. - Normal inflection points occur one standard deviation from the mean. - Match one- and two-standard-deviation central intervals with the Empirical Rule. - Use symmetry for the one-sided tail.

Solution

1. The density is symmetric and reaches its maximum at the filled marker \(x=4\), so \(\mu=4.\) 2. The open inflection markers are at \(x=3\) and \(x=5\). Since normal inflection points occur at \(\mu\pm\sigma\), \(\sigma=1.\) 3. The interval \([3,5]\) is \([\mu-\sigma,\mu+\sigma]\), so its probability is approximately \(68\%\). 4. The interval \([2,6]\) is \([\mu-2\sigma,\mu+2\sigma]\), so its probability is approximately \(95\%\). 5. The area from \(\mu\) to \(\mu+\sigma\) is about \(34\%\), so \(P(X\ge5)\approx50\%-34\%=16\%.\)

Answer

a) \(\mu=4\) and \(\sigma=1\). b) 1. Approximately \(68\%\) 2. Approximately \(95\%\) 3. Approximately \(16\%\)

53275212

The fill amount \(X\), in milliliters, of juice bottles is normally distributed. The graph shows its probability density function. The filled marker shows the peak, and the open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. Explain how the marked points support your values. b) Use the Empirical Rule to estimate the probability that a randomly selected bottle contains less than \(480\,\text{mL}\). Give the result as both a decimal and a percentage.

Figure for problem 532752

Hints

- Read the filled peak and open inflection markers rather than estimating an unmarked change in curvature. - For a normal density, inflection points occur at \(\mu\pm\sigma\). - Express \(480\) as a number of standard deviations below the mean. - Use symmetry to split the probability outside the central interval.

Solution

1. The filled peak is at \(x=500\), so \(\mu=500\,\text{mL}\). 2. The open inflection markers are at \(x=490\) and \(x=510\). Normal inflection points occur at \(\mu-\sigma\) and \(\mu+\sigma\), so \(\sigma=10\,\text{mL}\). 3. The cutoff is \(480=500-2(10)=\mu-2\sigma\). 4. By the Empirical Rule, approximately \(95\%\) of values lie within two standard deviations. The remaining \(5\%\) is split equally between the two tails, so \(P(X<480)\approx0.025\), or \(2.5\%\).

Answer

a) \(\mu=500\,\text{mL}\), \(\sigma=10\,\text{mL}\). b) Approximately \(0.025\), or \(2.5\%\).

53275612

The graph shows the density function \(f\) of a normally distributed random variable \(X\). The filled marker shows the peak, and the open markers show the inflection points. 1. Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. Briefly justify your answer. 2. Use a geometric estimate from the graph to show that \(P(5\le X\le7)>0.3.\) In particular, compare the curve on \([5,7]\) with the trapezoid formed by the endpoint heights.

Figure for problem 532756

Hints

- Read the peak and marked inflection points from the graph. - A normal density has inflection points at \(\mu\pm\sigma\). - On a concave-down interval, the chord joining two points lies below the curve. - Probability is area under the density curve.

Solution

1. The filled peak is at \(x=5\), so \(\mu=5.\) The open inflection markers are at \(x=3\) and \(x=7\). Since normal inflection points occur at \(\mu\pm\sigma\), \(\sigma=2.\) 2. On \([5,7]\), the density curve is concave down, so the chord joining the two endpoint points lies below the curve. 3. From the graph, \(f(5)\approx0.20,\qquad f(7)\approx0.12.\) The area of the trapezoid under the chord is approximately \(\frac{0.20+0.12}{2}(7-5)=0.32.\) Because this trapezoid lies below the density curve, \(P(5\le X\le7)>0.32>0.3.\)

Answer

1. \(\mu=5\) and \(\sigma=2\). 2. The trapezoid estimate is about \(0.32\), so \(P(5\le X\le7)>0.3\).

53276112

A juice company fills bottles labeled \(750\,\text{mL}\). The graph shows the approximately normal density of the actual fill amount \(X\). The filled marker shows the peak, and the open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. Then find the maximum density value to four decimal places. b) Use the Empirical Rule to estimate the probability that a randomly selected bottle: 1) contains less than \(750\,\text{mL}\); 2) contains between \(745\,\text{mL}\) and \(765\,\text{mL}\). c) The company lowers the mean to \(750\,\text{mL}\) while keeping the standard deviation unchanged. Describe how the density graph changes. Then estimate the percentage of bottles containing less than \(745\,\text{mL}\).

Figure for problem 532761

Hints

- Read the filled peak and open inflection markers from the graph. - Normal inflection points occur at \(\mu\pm\sigma\). - Express each cutoff in standard-deviation units. - Changing the mean shifts a normal curve horizontally without changing its spread.

Solution

1. The filled peak is at \(755\), so \(\mu=755\,\text{mL}\). The open inflection markers are at \(750\) and \(760\), so \(\sigma=5\,\text{mL}\). 2. The maximum density is \(\frac{1}{5\sqrt{2\pi}}\approx0.0798.\) 3. The cutoff \(750\) is one standard deviation below the mean. By the Empirical Rule, approximately \(16\%\) of bottles lie below it. 4. The interval \([745,765]\) is \([\mu-2\sigma,\mu+2\sigma]\), so its probability is approximately \(95\%\). 5. Lowering the mean from \(755\) to \(750\) shifts the whole density curve \(5\,\text{mL}\) to the left without changing its shape. 6. Under the new model, \(745=\mu-\sigma\), so approximately \(16\%\) of bottles contain less than \(745\,\text{mL}\).

Answer

a) \(\mu=755\,\text{mL}\), \(\sigma=5\,\text{mL}\), and the maximum density is approximately \(0.0798\). b) 1) Approximately \(16\%\). 2) Approximately \(95\%\). c) The curve shifts \(5\,\text{mL}\) left without changing shape; approximately \(16\%\) of bottles are below \(745\,\text{mL}\).

53276212

The fill amount \(X\), in milliliters, of apple juice bottles is normally distributed with mean \(\mu=1005\) and standard deviation \(\sigma=5\). A bottle is underfilled if it contains less than \(1000\,\text{mL}\). a) Use the Empirical Rule to estimate the probability that a randomly selected bottle is underfilled. b) The manufacturer is considering two ways to reduce the underfilled proportion: 1) Increase the mean while keeping the standard deviation fixed. 2) Decrease the standard deviation while keeping the mean fixed. Explain how each change affects the normal density and why each reduces the probability of underfilling.

Hints

- Express the underfill cutoff relative to the mean and standard deviation. - Use symmetry with the Empirical Rule. - Treat changes in center and changes in spread separately. - Keep the physical cutoff fixed while imagining how the density changes.

Solution

1. Since \(1000=1005-5=\mu-\sigma\), the underfill cutoff is one standard deviation below the mean. 2. By the Empirical Rule, approximately \(68\%\) lies within one standard deviation of the mean. The remaining \(32\%\) is split between the two tails, so the lower tail contains about \(16\%\). 3. Increasing the mean shifts the density curve to the right while keeping its shape unchanged. The fixed cutoff \(1000\) then lies farther into the left tail, so the underfilled proportion decreases. 4. Decreasing the standard deviation makes the density narrower and taller around the same mean. The fixed cutoff is then more standard deviations below the mean, so the underfilled proportion also decreases.

Answer

a) Approximately \(0.16\), or \(16\%\). b) Increasing the mean shifts the density right; decreasing the standard deviation makes it narrower and taller. Both changes move \(1000\,\text{mL}\) farther into the lower tail in standard-deviation terms, reducing the underfilled proportion.

53480612

A greenhouse studies the height \(X\), in centimeters, of young sunflowers. The heights are approximately normally distributed, and the graph shows the density function \(f\). The filled marker shows the peak, and the open markers show the inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. b) Estimate \(P(8\le X\le12)\) by using the grid squares under the curve. Each grid square has area \(0.1\). c) Without further calculation, determine \(P(X\ge10)\). d) Use your answers and the symmetry of the graph to estimate \(P(X>12)\).

Figure for problem 534806

Hints

- Read the peak and marked inflection points from the graph. - Probability is area under the density curve. - Use the stated grid-square area to estimate the central region. - Use symmetry about the mean to obtain the one-sided areas.

Solution

1. The filled peak is at \(x=10\), so \(\mu=10\,\text{cm}.\) The open inflection markers are at \(x=8\) and \(x=12\). Since normal inflection points occur at \(\mu\pm\sigma\), \(\sigma=2\,\text{cm}.\) 2. The area under the curve from \(8\) to \(12\) is about \(6.8\) grid squares. Since each square has area \(0.1\), \(P(8\le X\le12)\approx0.68.\) 3. By symmetry, half the total area lies at or above the mean: \(P(X\ge10)=0.5.\) 4. By symmetry, \(P(10\le X\le12)\approx0.68/2=0.34.\) Therefore, \(P(X>12)\approx0.50-0.34=0.16.\)

Answer

a) \(\mu=10\,\text{cm}\), \(\sigma=2\,\text{cm}\). b) Approximately \(0.68\). c) \(0.50\). d) Approximately \(0.16\).

53481012

The fill amount \(X\), in milliliters, of a manufacturer's juice bottles is approximately normally distributed. The graph shows the density function. The filled marker shows the peak, and the open markers show the inflection points. The target fill amount is \(1000\,\text{mL}\). Determine the mean \(\mu\) and standard deviation \(\sigma\) from the graph. Then find the probability that a randomly selected bottle contains more than \(1015\,\text{mL}\), rounded to four decimal places.

Figure for problem 534810

Hints

- Read the filled peak and open inflection markers. - Normal inflection points occur at \(\mu\pm\sigma\). - Convert the cutoff to a standard score. - Use an upper-tail probability.

Solution

1. The filled peak is at \(x=1000\), so \(\mu=1000\,\text{mL}\). 2. The open inflection markers are at \(990\) and \(1010\), so \(\sigma=10\,\text{mL}\). 3. The standard score for \(1015\) is \(z=(1015-1000)/10=1.5\). 4. Thus, \(P(X>1015)=1-\Phi(1.5)\approx0.0668\).

Answer

\(\mu=1000\,\text{mL}\), \(\sigma=10\,\text{mL}\), and \(P(X>1015)\approx0.0668\).

53481112

The length \(X\), in millimeters, of a manufactured part is normally distributed. The graph shows its density function. The filled marker shows the peak, and the open markers show the inflection points. A part meets specifications when its length is between \(37\,\text{mm}\) and \(43\,\text{mm}\). Determine \(\mu\) and \(\sigma\) from the graph, and find the proportion of parts that meet specifications. Round the proportion to four decimal places.

Figure for problem 534811

Hints

- Read the filled peak and open inflection markers. - Normal inflection points occur at \(\mu\pm\sigma\). - Standardize both specification limits. - Subtract cumulative probabilities to find the interval probability.

Solution

1. The filled peak is at \(x=40\), so \(\mu=40\,\text{mm}\). The open inflection markers are at \(38\) and \(42\), so \(\sigma=2\,\text{mm}\). 2. The endpoint standard scores are \(z_1=(37-40)/2=-1.5\) and \(z_2=(43-40)/2=1.5\). 3. Therefore, \(P(37\le X\le43)=\Phi(1.5)-\Phi(-1.5)\approx0.8664\).

Answer

\(\mu=40\,\text{mm}\), \(\sigma=2\,\text{mm}\), and the meeting-specifications proportion is approximately \(0.8664\).

53481212

The masses of oranges of a certain variety are approximately normally distributed with mean \(\mu=160\,\text{g}\) and standard deviation \(\sigma=20\,\text{g}\). a) Use the formula for the maximum of a normal density to show that the maximum density is about \(0.020\,\text{g}^{-1}\). b) Estimate the probability that a randomly selected orange has mass between \(140\,\text{g}\) and \(180\,\text{g}\). c) Suppose \(a\) is measured in grams. Interpret \(\int_{160}^{160+a}f(t)\,dt\approx0.341\) in context and find \(a\). d) Evaluate the criticism that a normal distribution theoretically allows negative masses.

Hints

- Recall the maximum-value formula for a normal density. - Identify the interval one standard deviation on either side of the mean. - An integral of a density function represents probability over an interval. - Express zero grams as a standard score to judge whether the model's impossible tail is practically important.

Solution

1. A normal density reaches its maximum at the mean, and the maximum value is \(1/(\sigma\sqrt{2\pi})\). Thus, \(1/(20\sqrt{2\pi})\approx0.01995\,\text{g}^{-1}\approx0.020\,\text{g}^{-1}\). 2. The interval \([140,180]\) is \([\mu-\sigma,\mu+\sigma]\), so \(P(140\le X\le180)\approx0.6827\), or about \(68.3\%\). 3. The integral is the probability that an orange has mass between \(160\,\text{g}\) and \((160+a)\,\text{g}\). By symmetry, the area from \(\mu\) to \(\mu+\sigma\) is about \(0.341\), so \(a=20\,\text{g}\). 4. The criticism is theoretically correct because the normal distribution has no lower bound. However, zero is eight standard deviations below the mean, so the probability of a negative mass is negligible; the normal model is therefore reasonable for this context.

Answer

a) \(1/(20\sqrt{2\pi})\approx0.020\,\text{g}^{-1}\). b) Approximately \(68.3\%\). c) The integral is the probability of a mass between \(160\,\text{g}\) and \(180\,\text{g}\); \(a=20\,\text{g}\). d) Negative values are theoretically possible under the model, but their probability is negligible.

53481312

The fill amount \(X\), in milliliters, of juice bottles labeled \(1\,\text{L}\) is normally distributed with mean \(\mu=1004\) and standard deviation \(\sigma=3\). a) Find the maximum value of the density function, rounded to three decimal places. b) Use symmetry and the Empirical Rule to estimate the probability that a randomly selected bottle contains less than \(1001\,\text{mL}\). c) Find \(k\) such that \(P(1004-k\le X\le1004+k)\approx0.95\). Explain what this statement means in context. d) Explain why a manufacturer generally sets the mean fill amount slightly above the labeled amount.

Hints

- The peak of a normal density occurs at the mean. - Express \(1001\) relative to the mean and standard deviation. - Recall the Empirical-Rule interval containing about \(95\%\) of values. - Consider what symmetry implies if the mean equals the labeled amount.

Solution

1. The maximum occurs at the mean and equals \(f(\mu)=1/(\sigma\sqrt{2\pi})=1/(3\sqrt{2\pi})\approx0.133\). 2. The cutoff \(1001=1004-3=\mu-\sigma\). By the Empirical Rule, about \(68\%\) lies within one standard deviation, leaving about \(32\%\) outside. Symmetry gives \(P(X<1001)\approx0.16\). 3. About \(95\%\) of a normal distribution lies within approximately two standard deviations of the mean. Thus, \(k\approx2\sigma=6\,\text{mL}\). 4. This means that about \(95\%\) of bottles contain between \(998\,\text{mL}\) and \(1010\,\text{mL}\). 5. Setting the mean above the labeled amount reduces the proportion of bottles that fall below the label claim. If the mean were exactly \(1000\,\text{mL}\), symmetry would put half of the modeled fill amounts below \(1000\,\text{mL}\).

Answer

a) Approximately \(0.133\). b) Approximately \(0.16\), or \(16\%\). c) \(k\approx6\,\text{mL}\); about \(95\%\) of bottles contain between \(998\,\text{mL}\) and \(1010\,\text{mL}\). d) A mean above the labeled amount lowers the modeled underfill proportion.

53483312

The graph shows the density function of a normally distributed random variable \(X\), representing the mass of a certain variety of apple in grams. a) Use the graph to determine the mean \(\mu\) and the maximum value of the density function. b) Use the maximum value to calculate the standard deviation \(\sigma\). Round to the nearest gram. c) Use the Empirical Rule to estimate \(P(140\le X\le160)\). d) Estimate \(P(X>170)\). Justify your answer using the parameters you found.

Figure for problem 534833

Hints

- The peak of the bell curve identifies the mean. - Relate the peak height to \(\sigma\) using the normal density formula. - Match each interval or cutoff to a number of standard deviations from the mean. - Use symmetry to divide the probability outside a centered interval between the two tails.

Solution

1. a) The peak occurs at \(x=150\), so \(\mu=150\,\text{g}\). The graph shows a maximum density of approximately \(0.040\). 2. b) For a normal density, \(f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}\). Thus, \(\sigma=\frac{1}{0.040\sqrt{2\pi}}\approx9.97\,\text{g}\), which rounds to \(10\,\text{g}\). 3. c) The interval \([140, 160]\) is \([\mu-\sigma, \mu+\sigma]\). By the Empirical Rule, its probability is about \(0.68\). 4. d) The cutoff is \(170=150+2\cdot10=\mu+2\sigma\). About \(95\%\) lies within two standard deviations, leaving \(5\%\) in both tails. By symmetry, the upper-tail probability is about \(0.025\).

Answer

a) \(\mu=150\,\text{g}\); maximum density approximately \(0.040\) b) \(\sigma\approx10\,\text{g}\) c) Approximately \(0.68\), or \(68\%\) d) Approximately \(0.025\), or \(2.5\%\)

52522812

The width of a precision component is normally distributed with standard deviation \(\sigma=0.05\,\text{mm}\). Find the maximum deviation \(d\) from the mean such that approximately \(95\%\) of components lie within the tolerance interval \([\mu-d,\mu+d].\) Report \(d\) to three decimal places.

Hints

- Translate the tolerance statement into a central probability interval. - Relate a symmetric central interval to the standard normal CDF. - Find the standard score for a central \(95\%\) interval. - Convert that standard score back to millimeters.

Solution

1. Set \(P(\mu-d\le X\le\mu+d)=0.95.\) 2. Standardizing gives \(P\left(-\frac{d}{\sigma}\le Z\le\frac{d}{\sigma}\right)=0.95.\) 3. Let \(z=\frac{d}{\sigma}.\) For a central \(95\%\) interval, \(2\Phi(z)-1=0.95,\) so \(\Phi(z)=0.975.\) 4. A table or inverse normal function gives \(z\approx1.96\). Therefore, \(d=z\sigma\approx1.96(0.05)=0.098\,\text{mm}.\)

Answer

\(d\approx0.098\,\text{mm}\).

52526612

A factory produces metal rods whose length \(L\), in centimeters, is normally distributed with mean \(\mu=200.5\) and standard deviation \(\sigma=0.8\). a) Find the percentage of rods shorter than \(199.0\,\text{cm}\), rounded to two decimal places. b) After calibration, the mean is exactly \(200.0\,\text{cm}\). Find the maximum standard deviation such that at least \(99\%\) of rods have lengths between \(198.0\,\text{cm}\) and \(202.0\,\text{cm}\). Round the allowable value down to two decimal places.

Hints

- Standardize the one-sided cutoff in part a). - Express the symmetric tolerance probability using \(\Phi\). - Find the standard-normal quantile associated with cumulative probability \(0.995\). - Follow the instruction to round downward rather than to the nearest hundredth.

Solution

1. For part a), \(z=\frac{199.0-200.5}{0.8}=-1.875.\) Thus, \(P(L<199.0)=\Phi(-1.875)\approx0.0304,\) or \(3.04\%\). 2. For part b), the tolerance interval is symmetric with radius \(2\,\text{cm}\). Require \(2\Phi\left(\frac{2}{\sigma}\right)-1\ge0.99.\) Therefore, \(\Phi\left(\frac{2}{\sigma}\right)\ge0.995.\) The \(0.995\) quantile is approximately \(2.5758\), so \(\sigma\le\frac{2}{2.5758}\approx0.7764\,\text{cm}.\) 3. Rounding downward to two decimal places gives \(0.77\,\text{cm},\) which safely satisfies the requirement.

Answer

a) \(3.04\%\). b) \(0.77\,\text{cm}\).

52527612

The fill weight \(X\) of flour packages produced by a machine is normally distributed with mean \(\mu=1000\,\text{g}\). A quality-control study finds that approximately \(95\%\) of packages have fill weights within \(12\,\text{g}\) of the mean. Find the standard deviation \(\sigma\) of the fill process. Round to two decimal places.

Hints

- Write the given condition as a symmetric interval about the mean. - Standardize the interval endpoints. - Determine the probability in each tail when \(95\%\) lies in the center. - Use the corresponding standard-normal critical value.

Solution

1. The statement describes the central interval \([1000-12,1000+12]=[988,1012],\) with probability \(0.95\). 2. Standardizing gives \(P\left(-\frac{12}{\sigma}\le Z\le\frac{12}{\sigma}\right)=0.95.\) 3. By symmetry, \(2\Phi\left(\frac{12}{\sigma}\right)-1=0.95,\) so \(\Phi\left(\frac{12}{\sigma}\right)=0.975.\) 4. The \(0.975\) standard-normal quantile is approximately \(1.96\). Thus, \(\frac{12}{\sigma}=1.96.\) 5. Therefore, \(\sigma=\frac{12}{1.96}\approx6.12\,\text{g}.\)

Answer

\(\sigma\approx6.12\,\text{g}\).

52532012

Let \(X\) be normally distributed with mean \(\mu\) and standard deviation \(\sigma\). a) Find, to four decimal places, the probability that a value of \(X\) is no more than \(1.5\) standard deviations from the mean. b) Show that for every normal random variable \(X\), the probability \(P(\mu-k\sigma\le X\le\mu+k\sigma)\) depends only on \(k\) and equals \(2\Phi(k)-1\), where \(\Phi\) is the standard normal cumulative distribution function.

Hints

- Translate “no more than \(1.5\) standard deviations from the mean” into a symmetric interval. - Standardize the general interval by subtracting \(\mu\) and dividing by \(\sigma\). - Use the symmetry identity \(\Phi(-z)=1-\Phi(z)\).

Solution

1. a) The event is \(\mu-1.5\sigma\le X\le\mu+1.5\sigma\). 2. Standardizing with \(Z=\frac{X-\mu}{\sigma}\) gives \(P(-1.5\le Z\le1.5)\). 3. Thus, \(P(-1.5\le Z\le1.5)=2\Phi(1.5)-1\approx2\cdot0.9332-1=0.8664\). 4. b) More generally, subtracting \(\mu\) and dividing by \(\sigma>0\) transforms the event into \(-k\le\frac{X-\mu}{\sigma}\le k\). 5. Since \(Z=\frac{X-\mu}{\sigma}\) is standard normal, the probability is \(P(-k\le Z\le k)=\Phi(k)-\Phi(-k)\). 6. By symmetry, \(\Phi(-k)=1-\Phi(k)\), so the probability equals \(2\Phi(k)-1\), which contains neither \(\mu\) nor \(\sigma\).

Answer

a) \(0.8664\) b) \(P(\mu-k\sigma\le X\le\mu+k\sigma)=2\Phi(k)-1\), so the probability depends only on \(k\).

52536012

Let \(Y\) be normally distributed with mean \(\mu\) and standard deviation \(\sigma\). a) Find \(P(\mu-\sigma\le Y\le\mu+\sigma)\), rounded to four decimal places, and explain why the result does not depend on the particular values of \(\mu\) and \(\sigma\). b) For the fixed interval \(I=[10,20]\) and a fixed value of \(\sigma\), determine the value of \(\mu\) that maximizes \(P(Y\in I)\). Explain your reasoning. c) Describe what happens to \(P(Y\le\mu)\) when \(\sigma\) is doubled.

Hints

- Standardize endpoints written as multiples of \(\sigma\) from \(\mu\). - Center a fixed interval on the peak of a symmetric density to maximize captured area. - Use symmetry to determine the probability to the left of the mean.

Solution

1. Standardizing the endpoints gives \(P(\mu-\sigma\le Y\le\mu+\sigma) =P(-1\le Z\le1) =2\Phi(1)-1 \approx0.6827.\) The parameters cancel during standardization, so this probability is the same for every normal distribution. 2. For fixed \(\sigma\), a normal density places the greatest area over a fixed-width interval when the interval is centered at the density peak. The midpoint of \([10,20]\) is \(15\), so the maximizing value is \(\mu=15.\) 3. Symmetry gives \(P(Y\le\mu)=0.5\) for every positive \(\sigma\). Doubling \(\sigma\) does not change this probability.

Answer

a) \(0.6827\); it depends only on the standardized endpoints \(-1\) and \(1\). b) \(\mu=15\). c) The probability remains \(0.5\).

52536512

A normal random variable \(X\) has standard deviation \(\sigma=100\). In each independent case, find all possible values of the mean \(\mu\) using the normal cumulative distribution function or inverse-normal technology. Round answers to the nearest tenth. a) \(P(900\le X\le1200)=0.4985\). b) \(P(X\le551)=0.025\).

Hints

- In part a, write the interval probability as a difference of two normal CDF values before solving for \(\mu\). - Use symmetry about the midpoint of the interval to anticipate two solutions. - In part b, convert the cumulative probability directly to a standard-normal quantile. - Keep full numerical precision until the final rounding.

Solution

1. For part a, solve \(\Phi((1200-\mu)/100)-\Phi((900-\mu)/100)=0.4985\). 2. The interval probability is symmetric as a function of \(\mu\) about the interval midpoint \(1050\), so two solutions are possible. 3. Numerical solution gives \(\mu\approx899.962\) or \(\mu\approx1200.038\), so to the nearest tenth the means are \(900.0\) and \(1200.0\). 4. For part b, \((551-\mu)/100=z_{0.025}\approx-1.959964\). 5. Therefore, \(\mu=551-100(-1.959964)\approx746.996\), so \(\mu\approx747.0\).

Answer

a) \(\mu\approx900.0\) or \(\mu\approx1200.0\). b) \(\mu\approx747.0\).

52538412

A normal random variable \(X\) has mean \(\mu=150\) and standard deviation \(\sigma=10\). Find the positive value \(c\) in each independent case. a) \(P(c\le X\le150)\approx0.4985\) b) \(P(X\le c)\approx0.025\) c) \(P(140\le X\le c)\approx0.8185\)

Hints

- Locate each region under a bell curve. - Use half of a central empirical-rule probability when an interval begins or ends at the mean. - Identify whether the boundary is above or below the mean. - Combine known areas on opposite sides of the mean.

Solution

1. a) The probability \(0.4985\) is approximately the area from \(\mu-3\sigma\) to \(\mu\). Therefore, \(c=150-3\cdot10=120\). 2. b) A left-tail probability of \(0.025\) corresponds to \(z\approx-1.96\). Thus, \(c=150-1.96\cdot10=130.4\). 3. c) Since \(140=\mu-\sigma\), the area from \(140\) to \(150\) is approximately \(0.3415\). To reach \(0.8185\), the area from \(150\) to \(c\) must be approximately \(0.477\), the area from the mean to two standard deviations above it. 4. Therefore, \(c=150+2\cdot10=170\).

Answer

a) \(c=120\) b) \(c=130.4\) c) \(c=170\)

52539512

The thickness \(X\), in millimeters, of glass panels from a manufacturer is normally distributed with mean \(\mu=6.00\) and standard deviation \(\sigma=0.15\). a) Find the probability that a randomly selected panel is between \(5.80\,\text{mm}\) and \(6.20\,\text{mm}\) thick. Round to four decimal places. b) Find the probability that a panel's thickness rounds to exactly \(6.0\,\text{mm}\) to the nearest tenth. Round to four decimal places. c) Give an interval of width \(0.20\,\text{mm}\) that contains the greatest possible proportion of panel thicknesses. Explain your choice. d) Find, to the nearest hundredth of a millimeter, the thickness that \(95\%\) of the panels exceed.

Hints

- Standardize the interval endpoints. - For a rounding question, determine the original-value interval that rounds to the stated measurement. - A normal density is highest at its mean, so a fixed-width interval captures most probability when centered there. - “Exceeded by \(95\%\)” identifies a lower percentile.

Solution

1. For part a), the endpoint standard scores are approximately \(-1.3333\) and \(1.3333\), so \(P(5.80\le X\le6.20)\approx0.8176.\) 2. A thickness rounds to \(6.0\,\text{mm}\) when \(5.95\le X<6.05.\) Therefore, \(P(5.95\le X<6.05) =\Phi(0.3333)-\Phi(-0.3333) \approx0.2611.\) 3. For a fixed width, a normal distribution places the greatest probability in an interval centered at its mean. Therefore, the best interval is \([5.90\,\text{mm},6.10\,\text{mm}].\) 4. The value exceeded by \(95\%\) of panels is the fifth percentile: \(d=6.00+0.15z_{0.05} \approx6.00+0.15(-1.6449) \approx5.75\,\text{mm}.\)

Answer

a) \(0.8176\). b) \(0.2611\). c) \([5.90\,\text{mm},6.10\,\text{mm}]\), centered at the mean. d) \(5.75\,\text{mm}\).

53482212

The graph shows the cumulative distribution function \(F\) of a normally distributed random variable \(X\) with mean \(\mu=10\). a) Read \(F(7)\) and \(F(13)\) from the graph, and use them to find \(P(7\le X\le13)\). b) Explain why symmetry of the density curve about \(x=\mu\) implies that the cumulative distribution function has point symmetry about \((\mu, 0.5)\).

Figure for problem 534822

Hints

- Locate \(7\) and \(13\) on the horizontal axis and read the corresponding cumulative values. - An interval probability is a difference of cumulative probabilities. - The total area is \(1\), so complementary tail areas add to \(1\). - Compare equal distances to the left and right of the mean.

Solution

1. a) From the graph, \(F(7)\approx0.16\) and \(F(13)\approx0.84\). Therefore, \(P(7\le X\le13)=F(13)-F(7)\approx0.84-0.16=0.68\). 2. b) Symmetry of the density about \(x=\mu\) means the left-tail area below \(\mu-a\) equals the right-tail area above \(\mu+a\). Thus, \(F(\mu-a)=1-F(\mu+a)\). The two cumulative values are equally far from \(0.5\) in opposite directions, which gives point symmetry about \((\mu, 0.5)\).

Answer

a) \(F(7)\approx0.16\), \(F(13)\approx0.84\), and \(P(7\le X\le13)\approx0.68\) b) Density symmetry gives \(F(\mu-a)=1-F(\mu+a)\), so the cumulative distribution graph is point-symmetric about \((\mu, 0.5)\).

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