- 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.

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 \(70\).

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

- 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.

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.

Approximately \(68\%\).

- For a symmetric bell curve, identify the horizontal location of its peak. - Compare how widely each curve spreads on the horizontal axis. - Do not confuse the peak height with the size of the standard deviation.

1. Panel a) is centered at \(0\); panel b) is centered at \(2\), so b) is farther to the right. 2. Panel b) has a wider and lower peak than a), so b) has the larger standard deviation.

Panel b) has both the center farther right and the larger standard deviation.

- Read the boundary value of the shaded region on the horizontal axis. - A shaded area under a density curve represents an interval probability. - Use the symmetry of the bell curve about its center.

1. The shaded region is to the left of the center \(0\), so it represents \(X\le0\). 2. A Normal curve is symmetric about its mean. Half of its total area lies to the left of the center, so \(P(X\le0)=0.5\).

The event is \(X\le0\); its probability is \(0.5\).

- Think about the total area on each side of a Normal curve's center. - An interval has width, but a single point does not.

1. A Normal curve is symmetric about its mean, so the area above \(60\) is \(0.5\). 2. A Normal variable is continuous, so a single exact value has probability \(0\). 3. Statement a) is true and statement b) is false.

a) True. b) False: \(P(X=60)=0\).

- 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.

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 \(P(X=105)=0.\)

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

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

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

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

- 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.

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 \(P(Z\ge2)=1-\Phi(2)\approx0.0228.\)

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

- Find the probability outside the central interval. - Use symmetry to divide the remaining probability between the tails. - How is the area to the left of the upper boundary related to the right tail?

1. The probability outside the central interval is \(1-0.954=0.046\). 2. By symmetry, each tail contains half this probability: \(\frac{0.046}{2}=0.023\). 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 \(0.023\).

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

- 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).

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 \(P(X\ge150)=1-\Phi(-1)=\Phi(1)\approx0.8413\).

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

- 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.

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, \(P(X\le500)=0.5\).

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

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

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, \(P(X\ge500)=0.5\).

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

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

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 \(z=2\), \(P(H>58)\approx0.0228\).

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

- Use the symmetry of a normal density about its mean. - How far is each endpoint from the center, and where is its reflection? - Reflect both endpoints and then write them in increasing order.

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

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

- 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.

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

\(0.8664\).

- 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.

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 \(54.67\%\).

\(54.67\%\).

- 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.

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, \([15{,}000-3\cdot800, 15{,}000+3\cdot800]=[12{,}600, 17{,}400]\,\text{h}\).

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

- 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.

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, \(P(X>760)\approx\frac{0.05}{2}=0.025\).

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

- 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.

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.

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

- 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.

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 \(\sigma=2\).

\(\sigma=2\)

- 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 \(1\).

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, \(P(X>150)=0.5\).

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

- 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.

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, \(P(80\le X\le95)=\Phi(1.5)-\Phi(0)\approx0.4332\).

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

- 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.

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, \(P(X<96)=\Phi(-2)\approx0.0228\).

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

- 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.

1. For \(250\,\text{g}\), \(z=(250-252)/1.5=-1.3333\ldots\), so \(P(X<250)\approx0.0912\). 2. For the interval from \(251\,\text{g}\) to \(254\,\text{g}\), the endpoint standard scores are \(-0.6667\ldots\) and \(1.3333\ldots\). Thus, \(P(251\le X\le254)\approx0.6563\). 3. For \(255\,\text{g}\), \(z=(255-252)/1.5=2\), so \(P(X\ge255)=1-\Phi(2)\approx0.0228\).

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

- 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.

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 \(P(X>21{,}000)=1-\Phi(1.25)\approx0.1056\).

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

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

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

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

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

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

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

- 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.

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.

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

- 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.

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 \(P(X<490)\approx2.5\%\).

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

- 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.

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, \(P(2\le X\le4)\approx0.68\).

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

- 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.

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.

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.

- 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.

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 \([940,1060]\).

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

- 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.

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.

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

- 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.

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, \(P(62.5\le X\le67.5) > P(X<62.5)+P(X>67.5).\)

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

- 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.

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 \(253\,\text{g}\).

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

- 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.

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.

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

- Read the two boundaries of the shaded region. - Compare the distance from each boundary to the center with the stated standard deviation. - Use the matching band from the Empirical Rule.

1. The shaded interval runs from \(-1\) to \(1\), exactly one standard deviation on either side of the mean. 2. By the Empirical Rule, about \(68\%\) of values lie in this interval.

About \(68\%\).

- Compare the locations of the two centers before looking at the shading. - A curve with greater spread has more of its area far from the center. - Compare each right-tail region relative to the fixed cutoff.

1. Both distributions are centered at \(0\), but panel b) has a wider curve and a larger standard deviation. 2. The cutoff \(2\) is farther into the right tail of the narrower distribution in a). 3. Therefore, the shaded probability above \(2\) is larger for b).

b). Its wider distribution puts more probability above the same cutoff of \(2\).

- Think about the interval containing roughly \(95\%\) of the distribution. - Decide which end of that interval borders the longest checkout times. - Use the mean and standard deviation to locate that boundary.

1. By the Empirical Rule, about \(95\%\) of values lie within two standard deviations of the mean. 2. The remaining \(5\%\) is split equally between the two tails, so the longest \(2.5\%\) starts at two standard deviations above the mean. 3. The cutoff is \(8+2(2)=12\,\text{minutes}\).

About \(12\,\text{minutes}\).

- 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.

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, \(P(X=\mu)=0\).

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 \(x=\mu\).

- 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 \(\Phi\).

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.

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.

- 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.

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. \(P(|X-505|>5)=2\Phi(-2.5)\approx0.0124.\)

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

- 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.

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 \(P(110\le X\le120) =\Phi\left(\frac43\right)-\Phi\left(\frac23\right) \approx0.1613.\)

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

- 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.

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 \(13.24\%\).

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

- 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.

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 \(1\).

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.

- 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.

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, \(P(X>76)=1-\Phi(2)\approx0.0228.\)

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

- 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.

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, \(P(X=250)=0\).

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

- Translate “at most” and “at least” into the appropriate normal tail areas. - Use symmetry for the interval centered on the mean. - Keep standard scores exact until the final cumulative-probability calculation. - What probability can a continuous distribution assign to one exact value?

1. For \(140\), the standard score is \(z=-\frac{5}{6}\), 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 \(-\frac{5}{12}\) and \(\frac{5}{12}\), so \(P(145\le M\le155)\approx0.3231\). 4. Because \(M\) is continuous, \(P(M=150.0)=0\).

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

- Where must the mean lie if opposite normal tails have equal area? - The two tail boundaries give a useful midpoint. - Reflect the endpoints of the later intervals across the mean you found.

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

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

- Standardize the lifetime cutoff first. - Use the one-bulb probability as the success probability in the binomial count. - Keep the unrounded normal probability when evaluating the binomial probability.

1. The standard score is \(z=\frac{13{,}500-15{,}000}{1200}=-1.25.\) Therefore, \(p=P(X<13{,}500)=\Phi(-1.25)\approx0.1056.\) 2. Let \(Y\sim\operatorname{Bin}(200,p)\), where \(p=\Phi(-1.25)\). 3. \(P(Y\le30)=\sum_{k=0}^{30}\binom{200}{k}p^k(1-p)^{200-k}\approx0.9805.\)

1. \(\Phi(-1.25)\approx0.1056\). 2. \(P(Y\le30)\approx0.9805\).

- 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.

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, \(P(32\le X\le56)=\Phi(2)-\Phi(-1)\approx0.8186\).

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) \(P(32\le X\le56)\approx0.8186\).

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

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, \(\Phi(1.5)-\Phi(-1.5)=2\Phi(1.5)-1\approx0.8664\).

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) \(P(7\le X\le13)\approx0.8664\).

- 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.

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.

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

- 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.

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.

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

- 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.

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 \(0.6827\).

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 \(510\,\text{g}\).

- 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.

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 \(0.5763\).

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

- Standardize the one-cup cutoff. - Use the resulting probability as the binomial success probability. - Keep the unrounded one-cup probability in the second calculation.

1. The standardized cutoff is \(z=\frac{195-200}{5}=-1.\) Thus, \(p=P(X<195)=\Phi(-1)\approx0.1587.\) 2. Let \(Y\sim\operatorname{Bin}(50,p)\). 3. \(P(Y\le10)=\sum_{k=0}^{10}\binom{50}{k}p^k(1-p)^{50-k}\approx0.8406.\)

a) \(\Phi(-1)\approx0.1587\). b) \(P(Y\le10)\approx0.8406\).

- Convert the specification endpoints to z-scores. - Use the one-bolt fit probability as the binomial success probability. - Keep the unrounded normal probability for the shipment calculation.

1. The endpoints have standard scores \(-1\) and \(1\), so \(p=P(9.95\le X\le10.05)=\Phi(1)-\Phi(-1)\approx0.6827.\) 2. Let \(Y\sim\operatorname{Bin}(200,p)\). 3. \(P(Y\ge140)=1-P(Y\le139)\approx0.3291.\)

a) \(\Phi(1)-\Phi(-1)\approx0.6827\). b) \(P(Y\ge140)\approx0.3291\).

- 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.

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 \(\frac{0.95450-0.68269}{0.95450}\cdot100\% \approx28.48\%.\)

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 \(28.48\%\).

- Standardize the one-LED cutoff. - Use the resulting lifetime probability as the binomial success probability. - Keep the unrounded value of \(p\) for the second calculation.

1. The standardized cutoff is \(z=\frac{1410-1500}{60}=-1.5.\) Thus, \(p=P(X\ge1410)=\Phi(1.5)\approx0.9332.\) 2. Let \(Y\sim\operatorname{Bin}(40,p)\). 3. \(P(Y\ge38)=P(Y=38)+P(Y=39)+P(Y=40)\approx0.4947.\)

a) \(\Phi(1.5)\approx0.9332\). b) \(P(Y\ge38)\approx0.4947\).

- 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.

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 \(34.15\%+49.85\%=84.0\%\).

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

- 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.

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 \(68.3\%\): \((99.7\%-68.3\%)/2=15.7\%\).

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

- 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.

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 \(0.03\).

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

- 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.

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 \(100\%-68.3\%=31.7\%\).

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

- 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.

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 \(50\%-68.3\%/2=15.85\%\).

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

- Match each central percentage to the appropriate standard-normal cutoff or empirical-rule distance. - Use symmetry about the mean to locate interval endpoints. - For the left tail beyond three standard deviations, how is the outside area split?

1. A central \(90\%\) interval uses \(z^*\approx1.645\). Thus, \(450\pm1.645(40)=450\pm65.8\), giving \(I\approx[384.2, 515.8]\). 2. A central probability of approximately \(0.954\) corresponds to two standard deviations from the mean, so \(I=[450-2(40), 450+2(40)]=[370, 530]\). 3. Since \(330=450-3(40)=\mu-3\sigma\), the probability outside the central three-standard-deviation interval is approximately \(1-0.997=0.003\). 4. By symmetry, the left tail is approximately \(\frac{0.003}{2}=0.0015\).

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

- 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.

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, \(c=1500\).

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

- 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.

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 \(\sigma=3\).

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

- 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.

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 \(\sigma=10\).

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

- 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.

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 \(47.7\%+34.15\%=81.85\%\).

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

- 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.

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 \(\sigma=3\).

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

- 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.

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 \(\sigma=20\).

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

- Express the acceptable interval around its mean. - How much probability can lie in each tail when at least the target percentage lies in the middle? - Find the matching standard-normal cutoff and solve for the standard deviation. - Will ordinary rounding keep a strict maximum constraint satisfied?

1. The requirement is \(P(745\le X\le755)\ge0.98\). 2. By symmetry, \(2\Phi\left(\frac{5}{\sigma}\right)-1\ge0.98\), so \(\Phi\left(\frac{5}{\sigma}\right)\ge0.99\). 3. The standard-normal \(0.99\) quantile is \(z_{0.99}\approx2.326347874\). Thus, the exact maximum is \(\sigma_{\max}=\frac{5}{z_{0.99}}\approx2.149291624\,\text{mL}\). 4. Rounding to the nearest four decimal places would give \(2.1493\,\text{mL}\), which is slightly too large. Rounding downward gives the greatest allowable four-decimal value, \(2.1492\,\text{mL}\).

Exact threshold: \(\sigma_{\max}=\frac{5}{z_{0.99}}\approx2.149291624\,\text{mL}\). Greatest allowable value to four decimal places: \(2.1492\,\text{mL}\).

- Turn the maximum unacceptable percentage into a minimum acceptable percentage. - How far are the tolerance endpoints from the mean? - Use a symmetric normal probability to find the cutoff for \(\sigma\). - A maximum constraint may require rounding in a different direction from ordinary rounding.

1. At least \(99\%\) must be acceptable, so \(P(119.7\le L\le120.3)\ge0.99\). 2. The tolerance interval has radius \(0.3\,\text{mm}\). By symmetry, \(2\Phi\left(\frac{0.3}{\sigma}\right)-1\ge0.99\), hence \(\Phi\left(\frac{0.3}{\sigma}\right)\ge0.995\). 3. The exact bound is \(\sigma_{\max}=\frac{0.3}{z_{0.995}}\), with \(z_{0.995}\approx2.575829304\), so \(\sigma_{\max}\approx0.116467345\,\text{mm}\). 4. Rounding to the nearest fourth decimal would give \(0.1165\,\text{mm}\), which exceeds the limit. The greatest safe four-decimal value is \(0.1164\,\text{mm}\).

Exact threshold: \(\sigma_{\max}=\frac{0.3}{z_{0.995}}\approx0.116467345\,\text{mm}\). Greatest allowable value to four decimal places: \(0.1164\,\text{mm}\).

- How far are the two specification limits from the mean, in standard deviations? - When is a whole package being counted rather than an individual pin being measured? - A percentage cutoff may fall between whole numbers of pins. Which counts satisfy “at least”?

1. The specification limits are one standard deviation below and above the mean. 2. For one pin, \(p=P(119.5\le X\le120.5)=\Phi(1)-\Phi(-1)\approx0.6827\). 3. Let \(Y\sim\operatorname{Bin}(150,p)\) count pins that meet specifications. At least \(65\%\) of \(150\) means at least \(98\) pins because \(0.65\cdot150=97.5\). 4. Using the unrounded value of \(p\), \(P(Y\ge98)=\sum_{k=98}^{150}\binom{150}{k}p^{k}(1-p)^{150-k}\approx0.8059\).

a) \(\Phi(1)-\Phi(-1)\approx0.6827\). b) \(P(Y\ge98)\approx0.8059\).

- Standardize the \(505\,\text{g}\) cutoff. - Convert \(10\%\) of \(400\) to a count. - Use the unrounded one-bag probability in the binomial calculation.

1. The standardized cutoff is \(z=\frac{505-500}{4}=1.25.\) Thus, \(p=P(W>505)=1-\Phi(1.25)\approx0.1056.\) 2. Let \(X\sim\operatorname{Bin}(400,p)\). 3. At most \(10\%\) of \(400\) means \(X\le40\). 4. \(P(X\le40)=\sum_{k=0}^{40}\binom{400}{k}p^k(1-p)^{400-k}\approx0.3950.\)

a) \(1-\Phi(1.25)\approx0.1056\). b) \(P(X\le40)\approx0.3950\).

- Standardize the one-bottle cutoff. - Use the resulting probability as the binomial success probability. - Use a complement and logarithms for the minimum inspection count.

1. The standardized cutoff is \(z=\frac{750-754}{4}=-1,\) so \(p=P(X<750)=\Phi(-1)\approx0.1587.\) 2. Let \(Y\sim\operatorname{Bin}(12,p)\). Then \(P(Y\le1)=(1-p)^{12}+12p(1-p)^{11}\approx0.4105.\) 3. For \(n\) inspected bottles, \(P(\text{at least one underfilled})=1-(1-p)^n.\) Requiring this to be at least \(0.95\) gives \(n\ge\frac{\ln(0.05)}{\ln(1-p)}\approx17.34.\) The minimum integer is \(18\).

a) \(\Phi(-1)\approx0.1587\). b) \(P(Y\le1)\approx0.4105\). c) \(18\) bottles.

- Relate \(0.10\,\text{mm}\) to the given standard deviation. - Use exactly the instructed approximation \(p=0.05\) in parts b) and c). - For the large-counts check, solve both inequalities and use the stronger requirement.

1. The rejection cutoff is \(0.10\,\text{mm}=2\sigma\) from the mean. By the Empirical Rule, about \(95\%\) of values lie within \(2\sigma\), so about \(5\%\) lie outside. 2. Using the instructed approximation \(p=0.05\), let \(X\sim\operatorname{Bin}(50,0.05)\). 3. \(P(X=2)=\binom{50}{2}(0.05)^2(0.95)^{48}\approx0.2611.\) 4. The large-counts conditions are \(0.05n\ge10\) and \(0.95n\ge10\). 5. These require \(n\ge200\) and \(n\ge10.53\ldots\), respectively, so the minimum integer is \(200\). 6. At \(n=200\), the two counts are \(10\) and \(190\), so both conditions hold.

a) About \(5\%\). b) \(P(X=2)\approx0.2611\). c) \(n=200\); the large-counts values are \(10\) and \(190\).

- 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.

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 \(\sigma=10\).

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

- Express each boundary as a number of standard deviations from the mean. - Use symmetry to split the probability outside the central interval. - For “at least one,” use the complement event of selecting no Underweight bags. - Use logarithms to solve for a variable in an exponent.

1. a) The endpoints are \(490=500-2\cdot5=\mu-2\sigma\) and \(510=500+2\cdot5=\mu+2\sigma\). 2. By the Empirical Rule, \(P(490\le X<510)\approx0.95\). 3. The remaining probability is about \(0.05\), split equally between the tails. Thus, \(P(X\ge510)\approx0.025\), and the Underweight probability is also \(p\approx0.025\). 4. b) For \(n\) independent bags, the probability of at least one Underweight bag is \(1-(1-p)^n\). We need \(1-(0.975)^n\ge0.95\). 5. Therefore, \((0.975)^n\le0.05\), so \(n\ge\frac{\ln(0.05)}{\ln(0.975)}\approx118.33\). 6. The smallest whole number that works is \(119\).

a) Standard: approximately \(95\%\); Overweight: approximately \(2.5\%\) b) \(119\) bags

- 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.

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 \(100\%\).

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 \(100\%\).

- 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.

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 \(\mu=600\).

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

- 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.

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 \(\mu\approx209.2\).

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

- 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.

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 \(0.683\).

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

- 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.

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, \(P(|X-\mu|\ge2\sigma)\approx1-0.954=0.046\).

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

- 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.

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 \([993.76\,\text{g}, 1006.24\,\text{g}]\).

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

- Compare the given density with the general normal density to identify its center and spread. - Find the endpoints one standard deviation on either side of the mean. - Use the inverse standard-normal cumulative distribution for the last percentile.

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 \([\mu-\sigma, \mu+\sigma]=[494, 506]\). Thus, \(P(494\le X\le506)=\Phi(1)-\Phi(-1)\approx0.6827\). 3. Require \(\Phi\left(\frac{k-500}{6}\right)=0.90\). The standard-normal \(0.90\) quantile is approximately \(1.281551566\), so \(k=500+6(1.281551566)\approx507.69\,\text{mL}\).

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

- 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.

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

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

- 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.

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, \(\sigma\approx\frac{20}{1.9991}\approx10.00.\)

\(\sigma\approx10.00\).

- 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.

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, \(\mu\approx500-80(-1.6449)\approx631.6\).

\(\mu\approx631.6\).

- Which supplied critical value corresponds to each central percentage? - How far from the mean should the interval endpoints lie? - Check that the midpoint of each completed interval is \(\mu\).

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

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

- 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.

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 \(1-0.997=0.003\).

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

- Standardize each fill cutoff. - Use the below-\(492\) probability as the binomial success probability in part c). - Keep the unrounded normal probability when evaluating the binomial probability.

1. The z-scores for \(495\) and \(505\) are \(-1.25\) and \(1.25\), so \(P(495\le X\le505)=\Phi(1.25)-\Phi(-1.25)\approx0.7887.\) 2. For \(492\), \(z=-2\), so \(p=P(X<492)=\Phi(-2)\approx0.0228.\) 3. Let \(Y\sim\operatorname{Bin}(12,p)\). Then \(P(Y\le1)=(1-p)^{12}+12p(1-p)^{11}\approx0.9706.\)

a) \(0.7887\). b) \(0.0228\). c) \(0.9706\).

- Standardize the interval endpoints and the \(16\)-hour cutoff. - Use the long-life probability as the success probability for the count. - Keep the unrounded value of \(p\) in the binomial calculation.

1. The z-scores for \(12\) and \(15\) are \(-4/3\) and \(2/3\), so \(P(12\le X\le15)\approx0.6563.\) 2. For \(16\), \(z=4/3\), so \(p=P(X>16)\approx0.0912.\) 3. Let \(Y\sim\operatorname{Bin}(50,p)\). Then \(P(Y\ge5)=1-\sum_{k=0}^{4}\binom{50}{k}p^k(1-p)^{50-k}\approx0.4844.\)

a) \(0.6563\). b) \(0.0912\). c) \(0.4844\).

- 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.

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).

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

- 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 \(\sigma\).

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 \(15.0\).

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

- 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.

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 \(\mu\approx1010.3\,\text{g}.\)

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

- 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.

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, \(P(|X-6.2|>0.3)=1-[\Phi(2)-\Phi(-2)]\approx0.0455\).

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

- 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.

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, \(c=40+4(2.57583)\approx50.30\).

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

- 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.

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, \(m=1000+8(-2.3263)\approx981.39\,\text{mL}.\)

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

- 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.

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, \(r=z^*\sigma\approx1.96(8.746)\approx17.14\).

\(r\approx17.14\), giving the approximate continuous interval \([72.86, 107.14]\).

- 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.

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 \(95\%\) (more precisely \(95.4\%\)) within two standard deviations under the normal approximation.

1. \(\mu=120\), \(\sigma=\sqrt{84}\approx9.17\). 2. Approximately \([101.67, 138.33]\); integer values \(102\) through \(138\). 3. Approximately \(95\%\) by the Empirical Rule (\(95.4\%\) for the normal model).

- 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.

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 \(68\%\).

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

- 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.

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 \(297\).

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

- 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.

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.

a) Continuous approximation \([174.60,225.40]\); integer outcomes \(175\) through \(225\). b) Wider: capturing \(99\%\) rather than \(95\%\) of the normal distribution requires extending farther from the mean.

- 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.

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.

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

- 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.

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 \(114\).

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 \(114\).

- 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.

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.

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

- 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.

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 \(390\).

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 \(390\).

- 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.

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 \(4.6\%\).

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

- 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 \(\sigma\).

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 \(0.3\%\).

a) \(9\). b) Continuous range: \([72.36,107.64]\); integer counts \(73\) through \(107\). c) Approximately \(0.3\%\).

- 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.

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 \(99.7\%\).

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

- 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.

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 \(95.4\%\).

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

- Find the binomial mean and standard deviation. - Check the large-count condition. - Build the interval with radius \(3\sigma\). - Compare the observed count with the interval before applying the stated decision rule.

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 given three-standard-deviation rule, this count is inconsistent with the reported rate; outcomes outside three standard deviations have total normal-approximation probability about \(0.3\%\).

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 and is inconsistent under the stated rule.

- 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.

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.

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.

- 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.

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 \(P(X\ge5)\approx50\%-34\%=16\%.\)

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

- 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.

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 \(2.5\%\).

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

- 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.

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, \(P(5\le X\le7)>0.32>0.3.\)

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

- 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.

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 \(745\,\text{mL}\).

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 \(745\,\text{mL}\).

- 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.

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.

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.

- 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.

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, \(P(X>12)\approx0.50-0.34=0.16.\)

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

- 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.

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, \(P(X>1015)=1-\Phi(1.5)\approx0.0668\).

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

- 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.

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, \(P(37\le X\le43)=\Phi(1.5)-\Phi(-1.5)\approx0.8664\).

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

- Recall where a normal density reaches its maximum and how its height depends on the standard deviation. - Compare the given mass interval with one standard deviation on each side of the mean. - A region's area under a density curve gives the probability of values in that region. - Consider how many standard deviations below the mean zero grams lies.

1. A normal density reaches its maximum at the mean. Its maximum value is \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{20\sqrt{2\pi}}\approx0.01995\,\text{g}^{-1}\approx0.020\,\text{g}^{-1}\). 2. The interval from \(140\,\text{g}\) to \(180\,\text{g}\) extends one standard deviation on each side of the mean. Its probability is approximately \(0.6827\), or \(68.3\%\). 3. The area in part c) represents the probability that an orange's mass is between \(160\,\text{g}\) and \((160+a)\,\text{g}\). About \(0.341\) of the normal distribution lies between the mean and one standard deviation above it, so \(a\approx20\,\text{g}\) to the nearest gram. 4. A normal distribution allows negative values, but a mass of zero is eight standard deviations below the mean. Such a value has negligible probability under this model, so the concern does not make the normal approximation unreasonable here.

a) \(\frac{1}{20\sqrt{2\pi}}\approx0.020\,\text{g}^{-1}\). b) Approximately \(68.3\%\). c) The area is the probability that an orange weighs between \(160\,\text{g}\) and \(180\,\text{g}\), approximately; \(a\approx20\,\text{g}\). d) The model allows negative masses, but their probability is negligible for these parameters.

- The peak of a normal density occurs at its mean; think about the unit of density when the measured variable is in milliliters. - Compare \(1001\,\text{mL}\) with the mean and standard deviation. - Recall which Empirical-Rule interval contains about \(95\%\) of the values. - Consider what symmetry implies if the mean is equal to the labeled amount.

1. The density reaches its maximum at the mean. Its height is \(f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{3\sqrt{2\pi}}\approx0.133\,\text{mL}^{-1}\). 2. The cutoff \(1001\,\text{mL}\) is one standard deviation below the mean. The Empirical Rule places approximately \(68\%\) within one standard deviation, leaving about \(16\%\) in each tail. Thus, \(P(X<1001)\approx0.16\). 3. About \(95\%\) of values lie within roughly two standard deviations of the mean, so \(k\approx2\sigma=6\,\text{mL}\). 4. This means approximately \(95\%\) of the bottles have fill amounts between \(998\,\text{mL}\) and \(1010\,\text{mL}\). 5. Setting the mean above the label reduces the proportion of bottles with less than the labeled amount. If the mean equaled \(1000\,\text{mL}\), symmetry would put half of the modeled fill amounts below the label.

a) Approximately \(0.133\,\text{mL}^{-1}\). 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 label lowers the modeled proportion of underfilled bottles.

- 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.

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 \(0.025\).

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 \(2.5\%\)

- Standardize the \(495\,\text{g}\) cutoff. - Use the one-package underweight probability as the binomial success probability. - Use the complement of no underweight packages for part 3.

1. The standard score is \(-1.25\), so \(p=P(X<495)=\Phi(-1.25)\approx0.1056.\) 2. Let \(Y\sim\operatorname{Bin}(100,p)\). Then \(P(Y=10)=\binom{100}{10}p^{10}(1-p)^{90}\approx0.1296.\) 3. For \(n\) packages, \(P(Y\ge1)=1-(1-p)^n.\) Requiring this to exceed \(0.99\) gives \(n>\frac{\ln(0.01)}{\ln(1-p)}\approx41.24.\) The minimum integer is \(42\).

1. \(\Phi(-1.25)\approx0.1056\). 2. \(P(Y=10)\approx0.1296\). 3. \(42\) packages.

- What does a symmetric tolerance interval centered at the mean leave outside it? - How much area must be in each tail if the central interval contains the target percentage? - Convert the matching standard-normal cutoff back into width units.

1. The smallest qualifying half-width satisfies \(P(\mu-d\le X\le\mu+d)=0.95\). 2. Standardizing gives \(P\left(-\frac{d}{\sigma}\le Z\le\frac{d}{\sigma}\right)=0.95\). 3. Let \(z=\frac{d}{\sigma}\). Symmetry gives \(2\Phi(z)-1=0.95\), so \(\Phi(z)=0.975\). 4. The standard normal quantile is \(z\approx1.959964\), so \(d=z\sigma\approx0.0979982\,\text{mm}\), which rounds to \(0.098\,\text{mm}\).

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

- 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.

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.

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

- 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.

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, \(\sigma=\frac{12}{1.96}\approx6.12\,\text{g}.\)

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

- Convert the stated distance from the mean into a symmetric interval. - Standardize the general interval using the normal variable's mean and standard deviation. - Use symmetry of the standard normal distribution to simplify the two cumulative probabilities.

1. In part 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, \(P(-1.5\le Z\le1.5)=2\Phi(1.5)-1\approx0.8664\). 4. In part b), for \(k\ge0\) and \(\sigma>0\), subtracting \(\mu\) and dividing by \(\sigma\) transforms the interval to \(-k\le\frac{X-\mu}{\sigma}\le k\). 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 is \(2\Phi(k)-1\), which contains neither \(\mu\) nor \(\sigma\).

a) \(0.8664\). b) For \(k\ge0\), \(P(\mu-k\sigma\le X\le\mu+k\sigma)=2\Phi(k)-1\), so this probability depends only on \(k\).

- 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.

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.

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

- 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.

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 \(\mu\approx747.0\).

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

- A probability between two values can be written as the difference of their cumulative probabilities. - For part a), decide which side of the mean contains the unknown boundary. - For part c), include the area below the given lower endpoint before finding the cumulative probability at the unknown upper endpoint. - Use inverse-normal values and round only at the end.

1. In part a), \(\Phi(0)-\Phi((c-150)/10)=0.4985\), so \(\Phi((c-150)/10)=0.0015\). Thus, \(c=150+10\Phi^{-1}(0.0015)\approx120.3226\), or \(120.3\) to the nearest tenth. 2. In part b), \((c-150)/10=\Phi^{-1}(0.025)\approx-1.9600\). Therefore, \(c\approx130.4004\), or \(130.4\) to the nearest tenth. 3. In part c), the lower endpoint has standard score \(-1\). Therefore, \(\Phi((c-150)/10)=0.8185+\Phi(-1)\approx0.9771553\). 4. Inverting gives \(c=150+10\Phi^{-1}(0.9771553\ldots)\approx169.9825\), or \(170.0\) to the nearest tenth.

a) \(c\approx120.3\) b) \(c\approx130.4\) c) \(c\approx170.0\)

- 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.

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: \(d=6.00+0.15z_{0.05} \approx6.00+0.15(-1.6449) \approx5.75\,\text{mm}.\)

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

- 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.

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 \((\mu, 0.5)\).

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 \((\mu, 0.5)\).