Free Math Worksheets

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

a) \(P(X\le100)=0.5\) b) \(P(X\le110)\approx0.8413\) c) \(P(90\le X\le110)\approx0.6827\) d) \(P(X>120)\approx0.0228\) e) \(P(X=105)=0\)

- Standardize each endpoint to convert to the standard normal distribution. - Express each boundary in terms of the mean and standard deviation. - Use symmetry when evaluating a cumulative probability at a negative standard score. - Use a standard normal table or appropriate technology.

1. Standardize the endpoints: \(z_1=\frac{85-100}{15}=-1\) and \(z_2=\frac{120-100}{15}=\frac43\approx1.3333\). 2. Therefore, \(P(85\le X\le120)=\Phi\left(\frac43\right)-\Phi(-1)\). 3. Using a standard normal table or technology, \(\Phi\left(\frac43\right)\approx0.9088\) and \(\Phi(-1)\approx0.1587\). Thus, the probability is approximately \(0.9088-0.1587=0.7501\).

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

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

1. a) By symmetry, \(\Phi(-2)=1-\Phi(2)\approx1-0.9772=0.0228\). 2. b) \(\int_{-2}^{2}\phi(x)\,\text{d}x=\Phi(2)-\Phi(-2)\approx0.9772-0.0228=0.9544\). 3. c) The right-tail area is \(1-\Phi(2)\approx0.0228\).

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

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

1. The probability outside the central interval is \(1-0.954=0.046\). 2. By symmetry, each tail contains half of this probability: \(0.046\div2=0.023\). 3. a) Therefore, \(P(X\le\mu-2\sigma)\approx0.023\). 4. b) The cumulative probability through the upper endpoint is \(1-0.023=0.977\). 5. c) The right-tail probability is also approximately \(0.023\).

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

- Use the standardization formula. - Apply symmetry when a standard score is negative. - For part c, identify whether the requested area lies to the left or right of the boundary. - Compare each value with the mean before calculating.

1. Standardize using \(z=\frac{x-200}{50}\). 2. a) For \(275\), \(z=1.5\), so \(P(X\le275)=\Phi(1.5)\approx0.9332\). 3. b) The standard scores are \(-2\) and \(1\). Thus, \(P(100<X<250)=\Phi(1)-\Phi(-2)\approx0.8413-0.0228=0.8185\). 4. c) For \(150\), \(z=-1\). Therefore, \(P(X\ge150)=1-\Phi(-1)=\Phi(1)\approx0.8413\).

a) \(P(X\le275)\approx0.9332\) b) \(P(100<X<250)\approx0.8185\) c) \(P(X\ge150)\approx0.8413\)

- Use symmetry about the mean. - An interval probability is a difference of cumulative probabilities. - Values more than two standard deviations from the mean lie in the tails. - Half of a symmetric normal distribution lies on each side of its mean.

1. Standardize using \(Z=\frac{X-500}{20}\). 2. a) The standard scores are \(-1\) and \(1\), so \(P(480<X<520)=\Phi(1)-\Phi(-1)\approx0.6827\). 3. b) For \(550\), \(z=2.5\), so \(P(X>550)=1-\Phi(2.5)\approx0.0062\). 4. c) The standard scores are \(-2.5\) and \(-0.5\). Thus, \(P(450\le X\le490)=\Phi(-0.5)-\Phi(-2.5)\approx0.3085-0.0062=0.3023\). 5. d) By symmetry, \(P(X\le500)=0.5\).

a) \(P(480<X<520)\approx0.6827\) b) \(P(X>550)\approx0.0062\) c) \(P(450\le X\le490)\approx0.3023\) d) \(P(X\le500)=0.5\)

- Translate each inequality into a statement about package weight. - Convert each boundary to a standard score. - Use symmetry about the mean. - Find interval probabilities by subtracting cumulative probabilities.

1. a) The event means a package contains less than \(491\,\text{g}\). The standard score is \(z=\frac{491-500}{6}=-1.5\), so \(P(X<491)=\Phi(-1.5)\approx0.0668\). 2. b) The event means a package contains between \(494\,\text{g}\) and \(512\,\text{g}\). The standard scores are \(-1\) and \(2\), so \(P(494\le X\le512)=\Phi(2)-\Phi(-1)\approx0.8186\). 3. c) The event means a package contains at least \(500\,\text{g}\). By symmetry about the mean, \(P(X\ge500)=0.5\).

a) Less than \(491\,\text{g}\): approximately \(0.0668\), or \(6.68\%\) b) Between \(494\,\text{g}\) and \(512\,\text{g}\): approximately \(0.8186\), or \(81.86\%\) c) At least \(500\,\text{g}\): \(0.5\), or \(50\%\)

- Translate each inequality into words. - Express each boundary in standard deviation units from the mean. - Use symmetry when boundaries are equally distant from the mean.

1. a) The event means a boy is at most \(50\) inches tall. Since \(z=\frac{50-54}{2}=-2\), \(P(H\le50)=\Phi(-2)\approx0.0228\). 2. b) The event means a boy is between \(52\) and \(56\) inches tall. These values are one standard deviation below and above the mean, so \(P(52<H<56)=P(-1<Z<1)\approx0.6827\). 3. c) The event means a boy is taller than \(58\) inches. Since \(z=\frac{58-54}{2}=2\), \(P(H>58)=1-\Phi(2)\approx0.0228\).

a) At most \(50\) inches: approximately \(2.28\%\) b) Between \(52\) and \(56\) inches: approximately \(68.27\%\) c) Taller than \(58\) inches: approximately \(2.28\%\)

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

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

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

- 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. \([12{,}600\,\text{h}, 17{,}400\,\text{h}]\)

- 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) \(\sigma=3\)

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

a) \(P(X\le125)\approx0.1587\) b) \(P(150\le X\le200)\approx0.4772\) c) \(P(100\le X\le200)\approx0.9545\) d) \(P(X>150)=0.5\)

- 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, \(P(60\le X\le70)=\Phi(-1)-\Phi(-2)\approx0.1587-0.0228=0.1359\). 2. For \([70, 80]\), the standard scores are \(-1\) and \(0\). Thus, \(P(70\le X\le80)=\Phi(0)-\Phi(-1)\approx0.5000-0.1587=0.3413\). 3. For \([80, 95]\), the standard scores are \(0\) and \(1.5\). Thus, \(P(80\le X\le95)=\Phi(1.5)-\Phi(0)\approx0.9332-0.5000=0.4332\).

1. \(P(60\le X\le70)\approx0.1359\), or \(13.59\%\) 2. \(P(70\le X\le80)\approx0.3413\), or \(34.13\%\) 3. \(P(80\le X\le95)\approx0.4332\), or \(43.32\%\)

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

1. a) The standard scores are \(-1.5\) and \(1.5\). Therefore, \(P(97\le X\le103)=\Phi(1.5)-\Phi(-1.5)\approx0.8664\). 2. b) The standard scores are \(1.5\) and \(2.5\). Therefore, \(P(103\le X\le105)=\Phi(2.5)-\Phi(1.5)\approx0.9938-0.9332=0.0606\). 3. c) For \(96\), \(z=-2\). Thus, \(P(X<96)=\Phi(-2)\approx0.0228\).

a) \(P(97\le X\le103)\approx0.8664\), or \(86.64\%\) b) \(P(103\le X\le105)\approx0.0606\), or \(6.06\%\) c) \(P(X<96)\approx0.0228\), or \(2.28\%\)

- Match the probability with a familiar central normal interval. - Express each endpoint as \(\mu\) plus or minus a multiple of \(\sigma\). - Check that the interval is centered at the mean.

1. The interval \([130, 170]\) is symmetric about \(150\) and has radius \(20\). 2. A central probability of approximately \(0.9544\) corresponds to an interval extending two standard deviations from the mean. 3. Therefore, \(2\sigma=20\), so \(\sigma=10\).

\(\sigma=10\)

- Identify the mean and standard deviation. - 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. a) The standard score is \(z=\frac{250-252}{1.5}=-1.3333\). Thus, \(P(X<250)=\Phi(-1.3333)\approx0.0912\). 2. b) The endpoint standard scores are \(z_1=\frac{251-252}{1.5}=-0.6667\) and \(z_2=\frac{254-252}{1.5}=1.3333\). 3. Therefore, \(P(251\le X\le254)=\Phi(1.3333)-\Phi(-0.6667)\approx0.6563\). 4. c) The standard score is \(z=\frac{255-252}{1.5}=2\). Thus, \(P(X\ge255)=1-\Phi(2)\approx0.0228\).

a) Approximately \(0.0912\), or \(9.12\%\) b) Approximately \(0.6563\), or \(65.63\%\) c) Approximately \(0.0228\), or \(2.28\%\)

- Identify the mean and standard deviation from the problem. - 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. a) The standard score is \(z=\frac{18800-20000}{800}=-1.5\). Thus, \(P(X\le18{,}800)=\Phi(-1.5)\approx0.0668\). 2. b) The endpoint standard scores are \(z_1=\frac{19600-20000}{800}=-0.5\) and \(z_2=\frac{21200-20000}{800}=1.5\). 3. Therefore, \(P(19{,}600\le X\le21{,}200)=\Phi(1.5)-\Phi(-0.5)\approx0.6247\). 4. c) The standard score is \(z=\frac{21000-20000}{800}=1.25\). Thus, \(P(X>21{,}000)=1-\Phi(1.25)\approx0.1056\).

a) Approximately \(0.0668\), or \(6.68\%\) b) Approximately \(0.6247\), or \(62.47\%\) c) Approximately \(0.1056\), or \(10.56\%\)

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

1. Standardize \(75\): \(z=\frac{75-60}{12}=1.25\). 2. Use the complement: \(P(X\ge75)=1-\Phi(1.25)\). 3. Since \(\Phi(1.25)\approx0.8944\), \(P(X\ge75)\approx1-0.8944=0.1056\).

\(P(X\ge75)\approx0.1056\), or \(10.56\%\)

- An interval probability is a difference of cumulative probabilities. - Standardize each endpoint. - Use symmetry if your table lists only positive standard scores.

1. Standardize the endpoints: \(z_1=\frac{10-12}{2.5}=-0.8\) and \(z_2=\frac{15-12}{2.5}=1.2\). 2. Therefore, \(P(10\le X\le15)=\Phi(1.2)-\Phi(-0.8)\). 3. Using \(\Phi(1.2)\approx0.8849\) and \(\Phi(-0.8)\approx0.2119\), the probability is \(0.8849-0.2119=0.6730\).

\(P(10\le X\le15)\approx0.6730\), or \(67.30\%\)

- Recall the empirical-rule percentages. - Multiply an event probability by the sample size to find an expected count. - Check whether the question asks for values inside or outside an interval.

1. a) The empirical rule gives a probability of approximately \(0.683\) within one standard deviation. The expected count is \(400\cdot0.683=273.2\), or about \(273\) components. 2. b) The probability outside two standard deviations is approximately \(1-0.954=0.046\). The expected count is \(400\cdot0.046=18.4\), or about \(18\) components. 3. c) The probability within three standard deviations is approximately \(0.997\). The expected count is \(400\cdot0.997=398.8\), or about \(399\) components.

a) About \(273\) components b) About \(18\) components c) About \(399\) components

- The mean is the \(x\)-coordinate of the maximum. - The inflection points are one standard deviation from the mean. - Use symmetry and the fact that the total area is \(1\). - Recall the empirical-rule percentage within one standard deviation of the mean.

1. a) The maximum occurs at \(x=3\), so \(\mu=3\). The inflection points are at \(x=2\) and \(x=4\), each one unit from the mean, so \(\sigma=1\). 2. b) By symmetry, half of the total area lies to the left of the mean. Therefore, \(P(X\le3)=0.5\). 3. c) The interval \([2, 4]\) is \([\mu-\sigma, \mu+\sigma]\). By the empirical rule, \(P(2\le X\le4)\approx0.683\).

a) \(\mu=3\), \(\sigma=1\) b) \(P(X\le3)=0.5\) c) \(P(2\le X\le4)\approx0.683\), or about \(68.3\%\)

- The center and peak of a normal density identify the mean. - Compare the widths of the two bell curves. - Inflection points occur one standard deviation from the mean.

1. a) Both density curves peak at \(x=500\), so both machines have mean \(\mu=500\,\text{g}\). 2. For \(f_1\), the inflection points are approximately \(500\pm2\), so \(\sigma_1=2\,\text{g}\). 3. For \(f_2\), the inflection points are approximately \(500\pm5\), so \(\sigma_2=5\,\text{g}\). 4. b) Machine 1 is more precise because its smaller standard deviation means that 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 because it has the smaller standard deviation.

- Use the symmetry of the normal density. - Express each endpoint as a number of standard deviations from the mean. - Recall the \(68\)-\(95\)-\(99.7\) Empirical Rule. - Split the probability outside a centered interval equally between the two tails.

1. a) A normal distribution is symmetric about its mean, so \(50\%\) of the loaves weigh more than \(1000\,\text{g}\). 2. b) The interval is \([\mu-\sigma, \mu+\sigma]=[980, 1020]\). By the Empirical Rule, its probability is about \(0.68\). 3. c) The cutoff is \(960=1000-2\cdot20=\mu-2\sigma\). About \(95\%\) lies within two standard deviations, leaving \(5\%\) in the two tails. Symmetry gives about \(2.5\%\) in the lower tail. 4. d) About \(99.7\%\) lies within three standard deviations. Since \(3\sigma=60\,\text{g}\), the interval is \([940\,\text{g}, 1060\,\text{g}]\).

a) \(50\%\) b) Approximately \(0.68\), or \(68\%\) c) Approximately \(2.5\%\) d) \([940\,\text{g}, 1060\,\text{g}]\)

- Probability is represented by area under the density curve. - Compare each interval’s location with the maximum. - The intervals have the same width. - The density decreases as you move away from the mean on the right side.

1. Each probability is an area under the density curve, and both intervals have width \(5\). 2. The density reaches its maximum at \(x=10\) and decreases for \(x>10\). Therefore, the density values over \([10, 15]\) are greater than the corresponding values farther to the right over \([15, 20]\). 3. Hence, \(P(10\le X\le15)>P(15\le X\le20)\).

\(P(10\le X\le15)\) is greater because the interval is closer to the mean, where the density is higher.

- Locate the x-coordinate directly below the peak of the bell curve. - Relate the peak and symmetry of a normal density to its mean. - Interpret interval probabilities as areas under the curve. - Recall the approximate probability within one standard deviation of the mean.

1. a) The density reaches its maximum at \(x=65\), so \(\mu=65\) inches. A normal density is symmetric about its mean, and its highest density occurs there. 2. b) The interval \([62.5, 67.5]\) is \([\mu-\sigma, \mu+\sigma]\). 3. By the empirical rule, the central interval contains approximately \(68.3\%\) of the probability. 4. The two outer regions contain the remaining \(100\%-68.3\%=31.7\%\). 5. Therefore, the central probability is greater than the sum of the two tail probabilities.

a) \(\mu=65\) inches; the normal density is symmetric and has its maximum at the mean. b) \(P(62.5\le X\le67.5)\) is greater: approximately \(68.3\%\) compared with \(31.7\%\) 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 \(253\,\text{g}\)

- 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 \(x=\mu\).

- What area lies under a density curve above a single 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. For a continuous random variable, the probability of any single value is \(P(X=x)=\int_x^x f(t)\,dt=0\), because an interval consisting of one point has width \(0\). 2. A weight rounds to \(500\,\text{g}\) when \(499.5\le X<500.5\). Standardizing gives \(P(499.5\le X<500.5)=\Phi(0.125)-\Phi(-0.125)\approx 0.0995\), or about \(9.95\%\). 3. A weight rounds to \(500.0\,\text{g}\) when \(499.95\le X<500.05\). Thus, \(P(499.95\le X<500.05)=\Phi(0.0125)-\Phi(-0.0125)\approx 0.0100\), or about \(1.00\%\). The rounding interval is one-tenth as wide as in part 2, so the probability is much smaller.

1. \(P(X=500)=0\) 2. \(P(499.5\le X<500.5)\approx 0.0995\) 3. \(P(499.95\le X<500.05)\approx 0.0100\). 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. - Interpret distances such as \(2\sigma\) and \(2.5\sigma\) from the mean. - For a deviation greater than a given amount, account for both tails.

1. a) Since \(505\) is the mean, \(P(X\le505)=0.5\). 2. b) The interval \([501, 509]\) is \([\mu-2\sigma, \mu+2\sigma]\). Thus, \(P(501\le X\le509)=\Phi(2)-\Phi(-2)\approx0.9545\). 3. c) For \(x=500\), \(z=\frac{500-505}{2}=-2.5\). Therefore, \(P(X<500)=\Phi(-2.5)\approx0.0062\). 4. d) Because \(X\) is continuous, \(P(X=505.0)=0\). 5. e) \(P(|X-505|>5)=P(X<500)+P(X>510)\). By symmetry, this is \(2\Phi(-2.5)\approx0.0124\).

a) \(P(X\le505)=0.5\) b) \(P(501\le X\le509)\approx0.9545\) c) \(P(X<500)\approx0.0062\) d) \(P(X=505.0)=0\) e) \(P(|X-505|>5)\approx0.0124\)

- Standardize each value before using the standard normal cumulative distribution function. - Use symmetry to evaluate negative standard scores and tail probabilities. - The empirical rule can help you check whether results near 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 \(\frac23\) and \(\frac43\). Therefore, \(P(110\le X\le120)=\Phi\left(\frac43\right)-\Phi\left(\frac23\right)\approx0.9088-0.7475=0.1613\).

1. \(P(85\le X\le115)\approx0.6827\), or \(68.27\%\) 2. \(P(X\le70)\approx0.0228\), or \(2.28\%\) 3. \(P(X>130)\approx0.0228\), or \(2.28\%\) 4. \(P(110\le X\le120)\approx0.1613\), 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. - Interpret probability as area under the density curve.

1. a) Standardizing gives \(P(497\le X\le506)=P(-1\le Z\le2)=\Phi(2)-\Phi(-1)\approx0.8186\). 2. b) The standard score for \(495\) is \(z=\frac{495-500}{3}=-\frac{5}{3}\). Therefore, \(P(X<495)=\Phi\left(-\frac{5}{3}\right)\approx0.0478\), or \(4.78\%\). 3. c) Rounding to \(500\,\text{mL}\) corresponds to the interval \([499.5, 500.5)\). 4. Thus, \(P(499.5\le X<500.5)=\Phi\left(\frac{1}{6}\right)-\Phi\left(-\frac{1}{6}\right)\approx0.1324\).

a) Approximately \(0.8186\), or \(81.86\%\) b) Approximately \(4.78\%\) c) Approximately \(0.1324\), or \(13.24\%\)

- Use the mirror symmetry of the normal curve. - After subtracting the middle area from \(1\), how is the remaining area divided? - What does greater spread imply about the shape of a curve whose total area stays \(1\)?

1. a) 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, \(P(Z\le 0)=0.5\). 2. b) The area outside \([-1,1]\) is approximately \(1-0.6827=0.3173\). Symmetry divides this equally between the two tails, so \(P(Z\ge 1)\approx \frac{0.3173}{2}=0.15865\). 3. c) A larger standard deviation means greater spread. The normal curve becomes wider and flatter, and its maximum height at \(x=\mu\) decreases so that the total area remains \(1\).

a) \(P(Z\le 0)=0.5\) b) \(P(Z\ge 1)\approx 0.15865\) c) The curve becomes wider and flatter, with a lower maximum at the mean.

- Examine the shape of the frequency distribution. - Use a weighted mean because values occur with different frequencies. - Divide the difference by the target value when finding the signed percent deviation. - Use the sign to decide whether the sample mean is above or below the target.

1. a) The frequency distribution is unimodal and approximately symmetric about \(500\,\text{g}\), which is consistent with a bell-shaped normal model for small random production variation. 2. b) The weighted total is \(497\cdot3+498\cdot12+499\cdot25+500\cdot30+501\cdot18+502\cdot9+503\cdot3=49{,}987\,\text{g}\). 3. Therefore, \(\overline{x}=\frac{49{,}987}{100}=499.87\,\text{g}\). 4. c) The signed percent deviation is \(\frac{499.87-500}{500}\cdot100\%=-0.026\%\). The sample mean is \(0.026\%\) below the target.

a) The data are unimodal and approximately symmetric. b) \(\overline{x}=499.87\,\text{g}\) c) \(-0.026\%\), meaning \(0.026\%\) below the target

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

1. a) The frequencies peak near \(25.0\,\text{mm}\) and decrease approximately symmetrically toward both ends. This shape is consistent with a normal model for many small production effects. 2. b) The weighted total is \(24.8\cdot12+24.9\cdot45+25.0\cdot82+25.1\cdot48+25.2\cdot13=5000.5\,\text{mm}\). 3. Thus, \(\mu_{\text{emp}}=\frac{5000.5}{200}=25.0025\,\text{mm}\). 4. The empirical variance is \(\frac{1}{200}\sum f_i(x_i-25.0025)^2\approx0.00964375\,\text{mm}^2\), so \(s_{\text{emp}}\approx0.0982\,\text{mm}\). 5. c) The signed percent deviation is \(\frac{25.0025-25.0}{25.0}\cdot100\%=0.01\%\).

a) The distribution is approximately symmetric and bell-shaped. b) \(\mu_{\text{emp}}=25.0025\,\text{mm}\); \(s_{\text{emp}}\approx0.0982\,\text{mm}\) c) \(0.01\%\) above the theoretical 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 \(\sigma=3\) miles per hour. 3. For \(76\) miles per hour, \(z=\frac{76-70}{3}=2\). 4. Therefore, \(P(X>76)=1-\Phi(2)\approx0.0228\).

a) \(\mu=70\,\text{mph}\); \(\sigma=3\,\text{mph}\) b) Approximately \(0.0228\), or \(2.28\%\)

- Standardize each boundary. - Subtract cumulative probabilities for an interval. - Use symmetry for a two-tail deviation event. - Recall the probability of one exact value for a continuous random variable.

1. A: The standard score is \(z=\frac{245-250}{4}=-1.25\). Thus, \(P(X<245)=\Phi(-1.25)\approx0.1056\). 2. B: The standard scores are \(-0.5\) and \(1.25\). Therefore, \(P(248\le X\le255)=\Phi(1.25)-\Phi(-0.5)\approx0.5858\). 3. C: A deviation greater than \(6\,\text{g}\) corresponds to \(|Z|>1.5\). By symmetry, \(P(|X-250|>6)=2[1-\Phi(1.5)]\approx0.1336\). 4. D: Because \(X\) is continuous, \(P(X=250)=0\).

A: Approximately \(0.1056\), or \(10.56\%\) B: Approximately \(0.5858\), or \(58.58\%\) C: Approximately \(0.1336\), or \(13.36\%\) D: \(0\)

- Interpret “at most” and “at least” as cumulative or tail events. - Use symmetry for an interval centered at the mean. - Keep the exact standard scores until the final probability is evaluated. - Recall the point-probability property of continuous distributions.

1. A: The standard score is \(z=\frac{140-150}{12}=-\frac{5}{6}\). Thus, \(P(M\le140)=\Phi\left(-\frac{5}{6}\right)\approx0.2023\). 2. B: The standard score is \(z=\frac{165-150}{12}=1.25\). Thus, \(P(M\ge165)=1-\Phi(1.25)\approx0.1056\). 3. C: The endpoints have standard scores \(-\frac{5}{12}\) and \(\frac{5}{12}\). Therefore, \(P(145\le M\le155)=\Phi\left(\frac{5}{12}\right)-\Phi\left(-\frac{5}{12}\right)\approx0.3231\). 4. D: Since \(M\) is continuous, \(P(M=150.0)=0\).

A: Approximately \(0.2023\), or \(20.23\%\) B: Approximately \(0.1056\), or \(10.56\%\) C: Approximately \(0.3231\), or \(32.31\%\) D: \(0\)

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

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

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

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

1. a) Subtracting \(P(X\le40)\) gives only the probability over \([40, 56]\). The lower boundary of the requested interval is \(32\), not \(40\). 2. b) The correct setup is \(P(X\le56)-P(X\le32)\). Standardizing gives \(\Phi(2)-\Phi(-1)\approx0.9772-0.1587=0.8185\).

a) The student 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) \(P(32\le X\le56)\approx0.8185\)

- Recall the full standardization formula. - What role does the standard deviation play in a standard score? - Use symmetry about the mean.

1. a) The student subtracts the mean but does not divide by the standard deviation. The correct setup is \(\Phi\left(\frac{13-10}{2}\right)-\Phi\left(\frac{7-10}{2}\right)=\Phi(1.5)-\Phi(-1.5)\). 2. b) By symmetry, \(\Phi(1.5)-\Phi(-1.5)=2\Phi(1.5)-1\approx2\cdot0.9332-1=0.8664\).

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. - Use the empirical rule or the standard normal cumulative distribution function. - Compare the positions of the two normal curves for the two means. - Where is a normal density highest? - Consider how a horizontal shift changes the area over a fixed interval.

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

a) \(P(37\le X\le53)\approx0.6827\), or about \(68.27\%\) b) The probability becomes smaller.

- Standardize the endpoints for each distribution. - How does a larger standard deviation change a normal density curve? - The total area remains \(1\) even when the curve becomes wider. - Consider whether greater spread places more or less probability near the mean.

1. a) The standard scores are \(-0.5\) and \(0.5\). Thus, \(P(8\le X\le12)=\Phi(0.5)-\Phi(-0.5)\approx0.6915-0.3085=0.3830\). 2. b) For \(Y\), the standard scores are \(-0.25\) and \(0.25\). Therefore, \(P(8\le Y\le12)=\Phi(0.25)-\Phi(-0.25)\approx0.5987-0.4013=0.1974\). 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) \(P(8\le X\le12)\approx0.3830\) b) \(P(8\le Y\le12)\approx0.1974\), which is smaller because \(Y\) has greater spread.

- Express the maximum deviation as an absolute-value event or a symmetric interval. - Convert the deviation to standard deviation units. - Use symmetry about the mean. - Relate a central interval from \(-z\) to \(z\) to the cumulative distribution function.

1. The event is \(\mu-12\le X\le\mu+12\). 2. Standardizing gives \(-1.5\le Z\le1.5\), because \(\frac{12}{8}=1.5\). 3. Therefore, \(P(|X-\mu|\le12)=\Phi(1.5)-\Phi(-1.5)=2\Phi(1.5)-1\approx2\cdot0.9332-1=0.8664\).

Approximately \(0.8664\), or \(86.64\%\)

- Standardize the package-weight cutoff. - Use a binomial model for the number of packages meeting the condition. - A difference of cumulative distribution values gives an interval probability. - Compare the interval endpoints with the mean and standard deviation.

1. a) The standard score is \(z=\frac{485-500}{10}=-1.5\). Thus, \(P(X\le485)=\Phi(-1.5)\approx0.0668\). 2. b) Let \(K\sim\operatorname{Bin}(50,p)\), where \(p=\Phi(-1.5)\approx0.06681\). Then \(P(K\le3)=\sum_{k=0}^{3}\binom{50}{k}p^k(1-p)^{50-k}\approx0.5689\). 3. c) The expression equals \(P(490<X\le510)\). Standardizing gives \(\Phi(1)-\Phi(-1)\approx0.6827\). 4. It is the probability that a randomly selected package has a fill weight between \(490\,\text{g}\) and \(510\,\text{g}\).

a) Approximately \(0.0668\), or \(6.68\%\) b) Approximately \(0.5689\), or \(56.89\%\) c) Approximately \(0.6827\); it is the probability of a fill weight between \(490\,\text{g}\) and \(510\,\text{g}\).

- Use a complement for a “greater than” normal probability. - For “at least one,” use the complement of no components meeting the condition. - Interpret a difference of cumulative probabilities as an interval probability. - Express \(1800\) and \(2200\) relative to the mean.

1. a) The standard score is \(z=\frac{2400-2000}{250}=1.6\). Thus, \(P(X>2400)=1-\Phi(1.6)\approx0.0548\). 2. b) Let \(K\) count components lasting longer than \(2400\) hours. Then \(P(K\ge1)=1-P(K=0)=1-(1-0.054799)^{20}\approx0.6760\). 3. c) The expression is \(P(1800<X\le2200)\), the probability that a component lasts between \(1800\) and \(2200\) hours. 4. The standard scores are \(-0.8\) and \(0.8\), so the probability is \(\Phi(0.8)-\Phi(-0.8)\approx0.5763\).

a) Approximately \(0.0548\), or \(5.48\%\) b) Approximately \(0.6760\), or \(67.60\%\) c) Approximately \(0.5763\); it is the probability of a lifetime between \(1800\) and \(2200\) hours.

- Find interval probabilities from the normal cumulative distribution function. - Recall the probability of one exact value for a continuous random variable. - Interpret a density integral as an interval probability. - Compute percent change relative to the original probability.

1. a) For \(\sigma=10\), the first interval has standard scores \(-2\) and \(2\), so its probability is approximately \(0.95450\). 2. The second interval has standard scores \(-0.1\) and \(0.1\), so its probability is approximately \(0.07966\). 3. Since \(X\) is continuous, \(P(X=1000)=0\). 4. b) The integral equals \(P(999.5\le X<1000.5)=P(-0.05\le Z<0.05)\approx0.03988\). It is the probability that the measured weight lies in \([999.5, 1000.5)\,\text{g}\) and therefore rounds to \(1000\,\text{g}\) to the nearest gram. 5. c) With \(\sigma=5\), the probabilities for the first two intervals become approximately \(P(-4\le Z\le4)=0.99994\) and \(P(-0.2\le Z\le0.2)=0.15852\). 6. The relative increases are \(\frac{0.99994-0.95450}{0.95450}\cdot100\%\approx4.76\%\) and \(\frac{0.15852-0.07966}{0.07966}\cdot100\%\approx99.01\%\). 7. The point probability remains \(0\); its percent change is undefined because the original value is zero.

a) Approximately \(0.95450\), \(0.07966\), and \(0\), respectively b) Approximately \(0.03988\); it is the probability that the weight rounds to \(1000\,\text{g}\). c) The interval probabilities increase by approximately \(4.76\%\) and \(99.01\%\); the point probability remains \(0\), so a percent change is undefined.

- Convert tolerance endpoints to standard scores. - Interpret interval probability as area under the density curve. - Recall why one exact value has probability zero for a continuous variable. - Compare the old and new interval probabilities using percent decrease.

1. a) The tolerance interval \([9.96, 10.04]\) has standard scores \(-2\) and \(2\), so its probability is approximately \(0.95450\). 2. The interval \([9.99, 10.01]\) has standard scores \(-0.5\) and \(0.5\), so its probability is approximately \(0.38292\). 3. Since \(X\) is continuous, \(P(X=10.000)=0\). 4. b) The integral equals \(P(9.995\le X<10.005)=P(-0.25\le Z<0.25)\approx0.19741\). It is the probability that the diameter lies in \([9.995, 10.005)\,\text{mm}\) and therefore rounds to \(10.00\,\text{mm}\) to the nearest hundredth. 5. c) With \(\sigma=0.04\), the tolerance endpoints have standard scores \(-1\) and \(1\), so the new probability is approximately \(0.68269\). 6. The percent decrease is \(\frac{0.95450-0.68269}{0.95450}\cdot100\%\approx28.48\%\).

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

- Write the interval whose endpoints are the mean plus or minus \(0.75\) standard deviation. - Standardize the two endpoints. - Use symmetry to express the central probability in terms of \(\Phi(0.75)\). - Convert the final decimal probability to a percentage.

1. The interval within \(0.75\) standard deviation of the mean is \([\mu-0.75\sigma, \mu+0.75\sigma]\). 2. Standardize with \(Z=\frac{X-\mu}{\sigma}\). The interval becomes \([-0.75, 0.75]\). 3. Therefore, \(P(-0.75\le Z\le0.75)=\Phi(0.75)-\Phi(-0.75)=2\Phi(0.75)-1\). 4. Using \(\Phi(0.75)\approx0.77337\), the probability is \(2\cdot0.77337-1=0.54674\), or about \(54.67\%\).

Approximately \(54.67\%\)

- 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 \(84.0\%\)

- Remember that the total area under the density curve is \(100\%\). - Use a complement for a region outside a central interval. - Identify which central areas must be added or subtracted. - Use the fact that each side of the mean contains \(50\%\) of the probability.

1. a) Since \(56=\mu-2\sigma\) and \(104=\mu+2\sigma\), the interval is the central two-standard-deviation interval. Its probability is approximately \(95.4\%\). 2. b) Since \(116=\mu+3\sigma\), the total probability outside the central three-standard-deviation interval is \(100\%-99.7\%=0.3\%\). By symmetry, the right tail contains \(0.3\%\div2=0.15\%\). 3. c) The interval is from \(\mu-3\sigma\) to \(\mu-\sigma\). Its probability is half the difference between the central \(99.7\%\) and central \(68.3\%\): \(\frac{99.7\%-68.3\%}{2}=15.7\%\).

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. a) The endpoints have standard scores \(-1\) and \(1\), so \(P(501\le X\le507)=\Phi(1)-\Phi(-1)\approx0.6827\). 2. b) We need \(P(X<495)\le0.03\). Equivalently, \(\frac{495-\mu}{3}\le z_{0.03}\), where \(z_{0.03}\approx-1.8808\). 3. Thus, \(495-\mu\le3\cdot(-1.8808)\), so \(\mu\ge500.6424\). 4. The smallest value to the nearest tenth that satisfies the condition is \(\mu=500.7\,\text{g}\). Indeed, \(\mu=500.6\) gives a probability slightly above \(0.03\), while \(500.7\) gives a probability below \(0.03\).

a) Approximately \(0.6827\), or \(68.27\%\) b) \(\mu=500.7\,\text{g}\)

- Express each boundary in standard deviation units from the mean. - Use symmetry about the mean. - Use the fact that the total probability is \(100\%\). - A half of a central interval lies on each side of the mean. - Translate “differs from the mean” into an absolute-value event.

1. a) The interval is from \(\mu-\sigma\) to \(\mu+2\sigma\). The area from \(\mu-\sigma\) to \(\mu\) is \(68.3\%\div2=34.15\%\), and the area from \(\mu\) to \(\mu+2\sigma\) is \(95.4\%\div2=47.7\%\). The total is \(81.85\%\). 2. b) The value \(384\) is \(\mu-2\sigma\). The probability outside the central two-standard-deviation interval is \(100\%-95.4\%=4.6\%\), so the left tail contains \(4.6\%\div2=2.3\%\). 3. c) The event is \(|X-\mu|>\sigma\), the complement of the central one-standard-deviation interval. Its probability is \(100\%-68.3\%=31.7\%\).

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

- Locate each requested region on a bell curve. - Express each endpoint as a number of standard deviations from the mean. - Use the fact that \(50\%\) of the area lies on each side of the mean. - Add or subtract areas from the \(68\text{-}95\text{-}99.7\) rule.

1. a) The interval is \([\mu-2\sigma, \mu+2\sigma]\), so its probability is approximately \(95.4\%\). 2. b) The interval is from \(\mu+2\sigma\) to \(\mu+3\sigma\). The probability is \(99.7\%\div2-95.4\%\div2=2.15\%\). 3. c) The cutoff is \(\mu+\sigma\). The area to the right is \(50\%-68.3\%\div2=15.85\%\).

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

- Match each central probability with a number of standard deviations from the mean. - Use symmetry about the mean. - Relate a central interval to the probability in the two tails. - Determine how many standard deviations \(330\) is below the mean.

1. a) A central \(90\%\) interval uses \(z^*\approx1.645\). Thus, \(450\pm1.645\cdot40=450\pm65.8\), so \(I\approx[384.2, 515.8]\). 2. b) A central probability of approximately \(0.954\) corresponds to two standard deviations from the mean. Thus, \(I=[450-2\cdot40, 450+2\cdot40]=[370, 530]\). 3. c) Since \(330=450-3\cdot40=\mu-3\sigma\), the total probability outside the central three-standard-deviation interval is approximately \(1-0.997=0.003\). 4. By symmetry, the left tail contains half of that probability, so \(P(X\le330)\approx0.003\div2=0.0015\).

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

- 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. a) Since \(1400=\mu+2\sigma\), the right-tail probability is \((100\%-95.4\%)\div2=2.3\%\). 2. b) Since \(1100=\mu-\sigma\), the left-tail probability is \((100\%-68.3\%)\div2=15.85\%\). 3. c) The interval 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. The total is \(47.7\%+34.15\%=81.85\%\).

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

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

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

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

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

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

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

- 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 \(\sigma=10\)

- 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) \(\mu=600\)

- Use the fact that half the distribution lies on each side of the mean. - Match each cumulative probability with a standard score or empirical-rule boundary. - Rewrite a right-tail probability as a left-tail probability when useful. - Decide whether each boundary lies above or below the mean.

1. a) A cumulative probability of approximately \(0.1585\) occurs at \(\mu-\sigma\). Therefore, \(160=\mu-20\), so \(\mu=180\). 2. b) Using the empirical rule, a cumulative probability of approximately \(0.9985\) occurs at \(\mu+3\sigma\). Therefore, \(250=\mu+3\cdot20\), so \(\mu=190\). 3. c) Since \(P(X\ge170)\approx0.975\), the area to the left of \(170\) is approximately \(0.025\). This corresponds to \(z\approx-1.96\). 4. Thus, \(170=\mu-1.96\cdot20=\mu-39.2\), so \(\mu=209.2\).

a) \(\mu=180\) b) \(\mu=190\) c) \(\mu=209.2\)

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

1. 1) Compare the function with \(f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\). The center is \(\mu=500\), and \(2\sigma^2=72\), so \(\sigma=6\). 2. 2) The required interval is \([\mu-\sigma, \mu+\sigma]=[494, 506]\). Thus, \(P(494\le X\le506)=\Phi(1)-\Phi(-1)\approx0.6827\). 3. 3) We need \(\Phi\left(\frac{k-500}{6}\right)=0.90\). The \(0.90\) standard normal quantile is \(z\approx1.2816\). 4. Therefore, \(k=500+6\cdot1.2816\approx507.69\,\text{mL}\).

1) \(\mu=500\,\text{mL}\), \(\sigma=6\,\text{mL}\) 2) Approximately \(0.6827\), or \(68.27\%\) 3) \(k\approx507.69\,\text{mL}\)

- 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 \(z\approx-1.645\). 3. Standardizing gives \(-1.645\approx\frac{500-\mu}{80}\). 4. Therefore, \(500-\mu\approx-131.6\), so \(\mu\approx631.6\).

\(\mu\approx631.6\)

- Match each central probability with its standard normal critical value. - Use the form \([\mu-z^*\sigma, \mu+z^*\sigma]\). - Use a standard normal table or inverse normal function when needed.

1. a) A central probability of \(0.683\) corresponds to \(z^*=1\). Thus, \(I=[120-15, 120+15]=[105, 135]\). 2. b) A central probability of \(0.950\) uses \(z^*\approx1.96\). Thus, \(I=[120-1.96\cdot15, 120+1.96\cdot15]=[90.6, 149.4]\). 3. c) A central probability of \(0.990\) uses \(z^*\approx2.58\). Thus, \(I=[120-2.58\cdot15, 120+2.58\cdot15]=[81.3, 158.7]\).

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. a) The interval radius is \(0.0328\,\text{mm}\), and \(0.0328\div0.02=1.64\). A central interval extending \(1.64\) standard deviations from the mean contains approximately \(0.899\). 2. b) The interval radius is \(0.04\,\text{mm}\), and \(0.04\div0.02=2\). The central two-standard-deviation interval contains approximately \(0.954\). 3. c) The deviation is \(0.06\div0.02=3\) standard deviations. The probability outside the central three-standard-deviation interval is approximately \(1-0.997=0.003\).

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

- Relate each endpoint to the mean in standard deviation units. - Standardize the interval for part a. - A larger standard deviation makes a normal curve wider and lower. - A normal density reaches its maximum at the mean. - Consider how shifting the curve changes area over a fixed interval.

1. a) The standard scores are \(-1.6\) and \(1.6\). Thus, \(P(72\le X\le88)=\Phi(1.6)-\Phi(-1.6)=2\Phi(1.6)-1\approx2\cdot0.9452-1=0.8904\). 2. b) 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. c) 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 decreases relative to part a.

a) \(P(72\le X\le88)\approx0.8904\) b) The probability decreases. c) The probability decreases.

- Check whether the interval is symmetric about the mean. - Match \(0.683\) with a central empirical-rule interval. - For a cumulative probability of \(0.950\), find the associated positive standard score. - Write the boundary as \(\mu+z\sigma\).

1. a) The interval is centered at \(250\) and has radius \(30\). A central probability of approximately \(0.683\) corresponds to one standard deviation, so \(\sigma=30\). 2. b) A cumulative probability of \(0.950\) corresponds to \(z\approx1.64\). Thus, \(274.6=250+1.64\sigma\). 3. Solving gives \(1.64\sigma=24.6\), so \(\sigma=15\).

a) \(\sigma=30\) b) \(\sigma=15\)

- Standardize the labeled weight in part a). - Find the standard normal quantile for a left-tail probability of \(0.005\). - Rearrange the standard-score equation for the unknown parameter.

1. a) The standard score is \(z=\frac{1000-1006}{4}=-1.5\). Therefore, \(P(X<1000)=\Phi(-1.5)\approx0.0668\). 2. b) For a lower-tail probability of \(0.005\), the standard normal quantile is \(z_{0.005}\approx-2.5758\). 3. With \(\mu=1006\), \(\frac{1000-1006}{\sigma}=-2.5758\). Thus, \(\sigma=\frac{-6}{-2.5758}\approx2.33\,\text{g}\). 4. c) With \(\sigma=4\), \(\frac{1000-\mu}{4}=-2.5758\). Solving gives \(\mu=1000+4\cdot2.5758\approx1010.3\,\text{g}\).

a) Approximately \(0.0668\), or \(6.68\%\) 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. - Use symmetry to simplify a central probability or a two-tail probability. - Express each deviation in standard deviation units.

1. For \(6.0\le X\le6.4\), the standard scores are \(z_1=\frac{6.0-6.2}{0.15}=-\frac{4}{3}\) and \(z_2=\frac{6.4-6.2}{0.15}=\frac{4}{3}\). Thus, \(P(6.0\le X\le6.4)=P\left(-\frac{4}{3}\le Z\le\frac{4}{3}\right)\approx0.8176\). 2. A deviation greater than \(0.3\,\text{g}\) means \(X<5.9\) or \(X>6.5\). Since \(\frac{0.3}{0.15}=2\), this is the area outside \([-2, 2]\) on the standard normal scale. 3. Therefore, \(P(|X-6.2|>0.3)=1-P(-2\le Z\le2)\approx1-0.9545=0.0455\).

1) Approximately \(0.8176\), or \(81.76\%\) 2) Approximately \(0.0455\), or \(4.55\%\)

- Match each probability with a familiar central interval or tail. - Use symmetry about the mean. - For a one-sided probability, determine the remaining probability in the opposite tail. - Recall the critical values for common central probabilities.

1. a) A cumulative probability of approximately \(0.1585\) occurs at one standard deviation below the mean. Thus, \(c=40-4=36\). 2. b) Since \(32=40-2\cdot4=\mu-2\sigma\), a central probability of approximately \(0.954\) requires the upper endpoint \(c=\mu+2\sigma=48\). 3. c) A right-tail probability of \(0.005\) corresponds to a central \(99\%\) interval and \(z\approx2.58\). Thus, \(c=40+2.58\cdot4=50.32\).

a) \(c=36\) b) \(c=48\) c) \(c=50.32\)

- For a rounding question, determine the interval of original values that round to the stated value. - A normal density is highest at its mean. - “Exceeded by \(95\%\)” identifies a lower percentile. - Use a binomial model for the count in part e).

1. a) The endpoint standard scores are \(z_1=\frac{5.80-6.00}{0.15}=-1.3333\) and \(z_2=\frac{6.20-6.00}{0.15}=1.3333\). Thus, \(P(5.80\le X\le6.20)\approx0.8176\). 2. b) 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. c) For a fixed width, a normal distribution places the greatest probability in an interval centered at its mean. The interval is \([5.90\,\text{mm}, 6.10\,\text{mm}]\). 4. d) We need \(P(X>d)=0.95\), so \(d\) is the fifth percentile. Therefore, \(d=6.00+0.15z_{0.05}\approx6.00+0.15\cdot(-1.6449)\approx5.75\,\text{mm}\). 5. e) For one panel, \(p=P(X>5.85)=P(Z>-1)=\Phi(1)\approx0.8413\). 6. If \(Y\sim\operatorname{Binomial}(50,p)\), then \(P(Y\ge40)=\sum_{k=40}^{50}\binom{50}{k}p^k(1-p)^{50-k}\approx0.8406\).

a) Approximately \(0.8176\), or \(81.76\%\) b) Approximately \(0.2611\), or \(26.11\%\) c) \([5.90\,\text{mm}, 6.10\,\text{mm}]\), centered at the mean d) \(5.75\,\text{mm}\) e) Approximately \(0.8406\), or \(84.06\%\)

- Determine the interval of values that round to the stated whole number. - Center a fixed-width interval at the peak of the normal density. - Translate “meet or exceed” into an upper-tail probability. - Use the probability from part a) in a binomial model for part e).

1. a) The standard score is \(z=\frac{500-505}{4}=-1.25\). Thus, \(P(X<500)=\Phi(-1.25)\approx0.1056\). 2. b) A weight rounds to \(505\,\text{g}\) when \(504.5\le X<505.5\). Therefore, \(P(504.5\le X<505.5)=\Phi(0.125)-\Phi(-0.125)\approx0.0995\). 3. c) A normal density is highest at its mean, so the width-\(5.0\) interval with the greatest probability is centered at \(505.0\): \([502.5\,\text{g}, 507.5\,\text{g}]\). 4. d) We need \(P(X\ge d)=0.85\), so \(P(X<d)=0.15\). Thus, \(d=505+4z_{0.15}\approx505+4\cdot(-1.0364)\approx500.9\,\text{g}\). 5. e) Let \(Y\) be the number of packages below \(500\,\text{g}\). Then \(Y\sim\operatorname{Binomial}(25,p)\), where \(p\approx0.1056\). 6. Therefore, \(P(Y\le3)=\sum_{k=0}^{3}\binom{25}{k}p^k(1-p)^{25-k}\approx0.7318\).

a) Approximately \(0.1056\), or \(10.56\%\) b) Approximately \(0.0995\), or \(9.95\%\) c) \([502.5\,\text{g}, 507.5\,\text{g}]\) d) \(500.9\,\text{g}\) e) Approximately \(0.7318\), or \(73.18\%\)

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

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

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

- 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. - Find the standard normal value with \(0.01\) of the area to its left. - Convert the standard score back to the original units.

1. a) The value \(984\) is \(1000-2\cdot8=\mu-2\sigma\). By the Empirical Rule, about \(95\%\) of values lie within \(2\sigma\), leaving about \(5\%\) outside. Symmetry places about \(2.5\%\) in the lower tail. 2. b) A difference greater than \(12\,\text{mL}\) corresponds to \(|Z|>\frac{12}{8}=1.5\). Thus, \(P(|X-1000|>12)=2\cdot[1-\Phi(1.5)]\approx0.1336\). 3. c) We need \(P(X<m)=0.01\). The first-percentile standard normal value is \(z_{0.01}\approx-2.3263\). 4. Therefore, \(m=1000+8\cdot(-2.3263)\approx981.39\,\text{mL}\).

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

- Find the expected number of tablets above the target. - Compute the standard deviation of a binomial random variable. - Identify the critical value for the central \(95\%\) of a normal distribution. - Multiply that critical value by the standard deviation.

1. Let \(X\sim\operatorname{Bin}(600,0.15)\). Its mean is \(\mu=np=600\cdot0.15=90\). 2. Its standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{600\cdot0.15\cdot0.85}=\sqrt{76.5}\approx8.746\). The large-count condition is satisfied. 3. A central \(95\%\) normal interval uses \(z^*\approx1.96\). 4. Therefore, \(r=z^*\sigma\approx1.96\cdot8.746\approx17.14\).

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

- Use the formulas for the mean and standard deviation of a binomial random variable. - Remember that only integer counts are possible. - Match a two-standard-deviation interval with its empirical-rule probability.

1. Since \(X\sim\operatorname{Bin}(400,0.3)\), \(\mu=np=400\cdot0.3=120\), and \(\sigma=\sqrt{np(1-p)}=\sqrt{400\cdot0.3\cdot0.7}=\sqrt{84}\approx9.17\). 2. The radius is \(2\sigma\approx18.33\), so the continuous interval is approximately \([120-18.33, 120+18.33]=[101.67, 138.33]\). 3. The possible integer values are \(102, 103,\ldots,138\). 4. The large-count condition is satisfied, and the empirical rule estimates the central two-standard-deviation probability as approximately \(95.4\%\).

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

- Find the continuous one-standard-deviation interval and identify the integers it contains. - Extend the integer interval by \(0.5\) at each end for the continuity correction. - Use the standard normal cumulative distribution function for the corrected endpoints.

1. The mean is \(\mu=np=100\cdot0.4=40\), and the standard deviation is \(\sigma=\sqrt{100\cdot0.4\cdot0.6}=\sqrt{24}\approx4.899\). 2. The one-standard-deviation interval is approximately \([40-4.899, 40+4.899]=[35.101, 44.899]\). Its integer values are \(36\) through \(44\). 3. To approximate \(P(36\le X\le44)\), apply the continuity correction and use \(35.5\le Y\le44.5\), where \(Y\) is normal with mean \(40\) and standard deviation \(\sqrt{24}\). 4. The standard scores are \(z_1=\frac{35.5-40}{\sqrt{24}}\approx-0.919\) and \(z_2\approx0.919\). 5. Therefore, \(P(36\le X\le44)\approx\Phi(0.919)-\Phi(-0.919)\approx0.6417\), or \(64.2\%\). 6. 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 approximately \(64.2\%\), which is closer to \(68\%\).

- Use the binomial formulas for the mean and standard deviation. - Check both expected successes and expected failures. - Identify the integer values inside the calculated continuous interval. - Use cumulative binomial probabilities to find an interval probability.

1. Since \(X\sim\operatorname{Bin}(400,0.15)\), \(\mu=np=60\), and \(\sigma=\sqrt{np(1-p)}=\sqrt{51}\approx7.14\). The large-count condition is satisfied because \(np=60\) and \(n(1-p)=340\). 2. The radius is \(1.96\sigma\approx13.997\), so the continuous interval is approximately \([46.003, 73.997]\). The integer values inside it are \(47\) through \(73\). 3. Using the binomial distribution, \(P(47\le X\le73)=P(X\le73)-P(X\le46)\approx0.9417\). 4. The result, about \(94.17\%\), is close to the central normal probability of \(95\%\).

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

- Find the mean and standard deviation of each binomial distribution. - Use the critical values for central \(90\%\) and \(95\%\) normal intervals. - Build each interval in the form \(\mu\pm z^*\sigma\). - Check the large-count condition before using a normal approximation.

1. a) The mean is \(\mu=np=600\cdot0.20=120\), and the standard deviation is \(\sigma=\sqrt{600\cdot0.20\cdot0.80}=\sqrt{96}\approx9.798\). 2. For a central \(90\%\) interval, \(z^*\approx1.645\). Thus, \(120\pm1.645\cdot9.798\approx120\pm16.12\), giving \([103.88, 136.12]\). The integer outcomes inside this interval are \(104\) through \(136\). 3. b) The mean is \(\mu=400\cdot0.70=280\), and the standard deviation is \(\sigma=\sqrt{400\cdot0.70\cdot0.30}=\sqrt{84}\approx9.165\). 4. For a central \(95\%\) interval, \(z^*\approx1.96\). Thus, \(280\pm1.96\cdot9.165\approx280\pm17.96\), giving \([262.04, 297.96]\). The integer outcomes inside this interval are \(263\) through \(297\). 5. In both cases, \(np\) and \(n(1-p)\) are greater than \(10\), so the normal approximation is appropriate.

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. - Use the critical value for the central \(95\%\) of a normal distribution. - Consider how the width changes when an interval must contain more probability.

1. a) The mean is \(\mu=np=1250\cdot0.16=200\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{1250\cdot0.16\cdot0.84}=\sqrt{168}\approx12.961\). The large-count condition is satisfied. 3. For a central \(95\%\) interval, \(z^*\approx1.96\). Therefore, \(200\pm1.96\cdot12.961\approx200\pm25.40\), giving \([174.60, 225.40]\). 4. The integer outcomes inside the continuous interval are \(175\) through \(225\). 5. b) A central \(99\%\) interval is wider because it must include more probability, so it requires a larger critical value than the \(95\%\) interval.

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

- Find the binomial mean and standard deviation. - Use the critical value for a central \(95\%\) normal interval. - Convert the continuous interval to possible integer counts.

1. Let \(X\sim\operatorname{Bin}(1000,0.04)\). The mean is \(\mu=np=1000\cdot0.04=40\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{1000\cdot0.04\cdot0.96}=\sqrt{38.4}\approx6.197\). Since \(np=40\) and \(n(1-p)=960\), the normal approximation is appropriate. 3. A central \(95\%\) interval uses \(z^*\approx1.96\). 4. The radius is \(1.96\cdot6.197\approx12.15\), so the continuous interval is \(40\pm12.15\), or \([27.85, 52.15]\). 5. The integer counts inside this interval are \(28\) through \(52\).

\(28\) through \(52\) defective screws

- Interpret the binomial mean as an expected count. - Use the binomial standard deviation formula. - Build an interval symmetric about the mean. - Recall the empirical-rule percentage for three standard deviations.

1. a) Let \(X\sim\operatorname{Bin}(2400,0.05)\). The mean is \(\mu=np=2400\cdot0.05=120\). 2. b) The standard deviation is \(\sigma=\sqrt{2400\cdot0.05\cdot0.95}=\sqrt{114}\approx10.68\). 3. The radius is \(3\sigma\approx32.03\), so the continuous interval is approximately \([87.97, 152.03]\). The integer counts inside it are \(88\) through \(152\). 4. The large-count condition is satisfied because \(np=120\) and \(n(1-p)=2280\), 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) \(88\) through \(152\) packages; approximately \(99.7\%\)

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

1. a) Let \(X\sim\operatorname{Bin}(6000,0.015)\). The mean is \(\mu=np=6000\cdot0.015=90\). 2. The standard deviation is \(\sigma=\sqrt{6000\cdot0.015\cdot0.985}=\sqrt{88.65}\approx9.42\). The large-count condition is satisfied. 3. b) For a central \(95\%\) interval, use \(z^*\approx1.96\): \(90\pm1.96\cdot9.415\approx[71.55, 108.45]\). The integer counts are \(72\) through \(108\). 4. For a central \(99\%\) interval, use \(z^*\approx2.58\): \(90\pm2.58\cdot9.415\approx[65.71, 114.29]\). The integer counts are \(66\) through \(114\).

a) \(\mu=90\); \(\sigma\approx9.42\) b) \(95\%\): \(72\) through \(108\); \(99\%\): \(66\) 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.

- Identify the binomial model. - Find its mean and standard deviation. - Use the critical value for a central \(95\%\) interval. - Convert the continuous endpoints to possible integer counts.

1. Let \(X\sim\operatorname{Bin}(2500,0.04)\). The mean is \(\mu=np=2500\cdot0.04=100\). 2. The standard deviation is \(\sigma=\sqrt{2500\cdot0.04\cdot0.96}=\sqrt{96}\approx9.798\). The large-count condition is satisfied because \(np=100\) and \(n(1-p)=2400\). 3. A central \(95\%\) interval uses \(z^*\approx1.96\). Thus, \(100\pm1.96\cdot9.798\approx[80.80, 119.20]\). 4. The integer counts inside the interval are \(81\) through \(119\).

\(81\) through \(119\) defective components

- Find the binomial mean and standard deviation. - Recall the critical values for central \(90\%\) and \(99\%\) intervals. - Check the large-count condition. - Give integer counts inside each continuous interval.

1. Let \(X\sim\operatorname{Bin}(600,0.60)\). The mean is \(\mu=600\cdot0.60=360\), and the standard deviation is \(\sigma=\sqrt{600\cdot0.60\cdot0.40}=\sqrt{144}=12\). The large-count condition is satisfied because \(np=360\) and \(n(1-p)=240\). 2. For a central \(90\%\) interval, use \(z^*\approx1.64\): \(360\pm1.64\cdot12=[340.32, 379.68]\). The integer counts are \(341\) through \(379\). 3. For a central \(99\%\) interval, use \(z^*\approx2.58\): \(360\pm2.58\cdot12=[329.04, 390.96]\). The integer counts are \(330\) through \(390\).

Central \(90\%\) interval: \(341\) through \(379\) Central \(99\%\) interval: \(330\) through \(390\)

- Identify the binomial parameters. - Use the binomial mean and standard deviation formulas. - Match common central probabilities with their critical values. - A deviation from the mean can occur in either direction.

1. The mean is \(\mu=np=1600\cdot0.20=320\), and the standard deviation is \(\sigma=\sqrt{1600\cdot0.20\cdot0.80}=16\). The large-count condition is satisfied because \(np=320\) and \(n(1-p)=1280\), so a normal approximation is appropriate. 2. A central \(95\%\) interval uses \(z^*\approx1.96\). Thus, \(320\pm1.96\cdot16=[288.64, 351.36]\). The integer counts inside it are \(289\) through \(351\). 3. A deviation of \(32\) equals \(2\sigma\). The empirical rule places approximately \(95.4\%\) within two standard deviations, so the probability of a larger deviation is approximately \(100\%-95.4\%=4.6\%\).

1. \(\mu=320\); \(\sigma=16\) 2. \(289\) through \(351\) 3. Approximately \(4.6\%\)

- Find the mean and standard deviation first. - Use approximately \(68.3\%\), \(95.4\%\), and \(99.7\%\) for central one-, two-, and three-standard-deviation intervals. - Distinguish between probability inside and outside an interval. - For an exact central \(95\%\) approximation, use \(1.96\) rather than \(2\).

1. The mean is \(\mu=np=900\cdot0.10=90\), and the standard deviation is \(\sigma=\sqrt{900\cdot0.10\cdot0.90}=\sqrt{81}=9\). The large-count condition is satisfied because \(np=90\) and \(n(1-p)=810\). 2. a) The probability outside the central one-standard-deviation interval is approximately \(31.7\%\), so the requested deviation is \(9\). 3. b) A total of \(5\%\) outside corresponds to a central \(95\%\) interval. Using \(z^*\approx1.96\), the continuous interval is \(90\pm1.96\cdot9=[72.36, 107.64]\). The integer counts inside are \(73\) through \(107\). 4. c) Since \(27=3\sigma\), the probability of a deviation greater than \(27\) is approximately the probability outside the central three-standard-deviation interval, or \(0.3\%\).

a) \(9\) b) \(73\) to \(107\) c) \(0.3\%\)

- Identify the binomial model. - Find its mean and standard deviation. - Interpret one, two, and three standard deviations as interval radii. - Follow the instruction to round the endpoints outward.

1. Since \(X\sim\operatorname{Bin}\left(1800,\frac{1}{6}\right)\), \(\mu=np=300\), and \(\sigma=\sqrt{np(1-p)}=\sqrt{250}\approx15.811\). The large-count condition is satisfied because \(np=300\) and \(n(1-p)=1500\). 2. The one-standard-deviation interval is approximately \([300-15.811, 300+15.811]=[284.189, 315.811]\). Rounding outward gives \([284, 316]\). 3. The two-standard-deviation interval is approximately \([268.377, 331.623]\). Rounding outward gives \([268, 332]\). 4. The three-standard-deviation interval is approximately \([252.566, 347.434]\). Rounding outward gives \([252, 348]\).

1. \([284, 316]\) 2. \([268, 332]\) 3. \([252, 348]\)

- Check the expected numbers of successes and failures. - Match \(95.4\%\) with a number of standard deviations. - Remember that a count can take only integer values.

1. The mean is \(\mu=np=2000\cdot0.03=60\), and the standard deviation is \(\sigma=\sqrt{2000\cdot0.03\cdot0.97}=\sqrt{58.2}\approx7.629\). 2. a) The large-count condition is satisfied because \(np=60\) and \(n(1-p)=1940\), so a normal approximation is appropriate. 3. b) A central probability of approximately \(95.4\%\) corresponds to two standard deviations. The continuous interval is \(60\pm2\cdot7.629\approx[44.742, 75.258]\). 4. The integer counts inside the interval are \(45\) through \(75\).

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

- Find the expected number of patients with the reaction. - Build an interval with radius three standard deviations. - Compare the observed count with the interval. - Recall the approximate probability outside a three-standard-deviation interval.

1. Let \(X\sim\operatorname{Bin}(4000,0.025)\). The mean is \(\mu=4000\cdot0.025=100\). 2. The standard deviation is \(\sigma=\sqrt{4000\cdot0.025\cdot0.975}=\sqrt{97.5}\approx9.874\). The large-count condition is satisfied because \(np=100\) and \(n(1-p)=3900\). 3. a) The continuous three-standard-deviation interval is \(100\pm3\cdot9.874\approx[70.38, 129.62]\). The integer counts inside it are \(71\) through \(129\). 4. b) The observed count \(160\) is well above the interval. Results outside a three-standard-deviation interval have total probability about \(0.3\%\) under the model, so this result is not consistent with the reported rate by this criterion.

a) \(71\) through \(129\) patients b) No. The observed count \(160\) is outside the three-standard-deviation interval.

- The peak of a normal density identifies its mean. - The inflection points occur one standard deviation from the mean. - Compare the cutoff with each distribution's mean and use symmetry. - Substitute \(x=\mu\pm\sigma\) into the density formula.

1. a) The mean is the x-coordinate of the peak. Thus, \(\mu_A=10\,\text{g}\) and \(\mu_B=14\,\text{g}\). 2. The horizontal distance from the mean to either inflection point is the standard deviation. The inflection points of \(f_A\) occur at \(x=8\) and \(x=12\), so \(\sigma_A=2\,\text{g}\). Those of \(f_B\) occur at \(x=13\) and \(x=15\), so \(\sigma_B=1\,\text{g}\). 3. b) For Machine A, \(12\,\text{g}\) is above the mean, so more than half of the distribution lies below \(12\,\text{g}\). For Machine B, \(12\,\text{g}\) is below the mean, so less than half lies below \(12\,\text{g}\). Therefore, Machine A produces the greater underfilled proportion. 4. c) For Machine B, \(f_B(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-14)^2}\). At \(x=14\pm1\), the y-coordinate is \(\frac{1}{\sqrt{2\pi}}e^{-1/2}=\frac{1}{\sqrt{2\pi e}}\). 5. The inflection points are \(\left(13, \frac{1}{\sqrt{2\pi e}}\right)\) and \(\left(15, \frac{1}{\sqrt{2\pi e}}\right)\), with y-coordinate approximately \(0.2420\).

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) Machine A produces a greater proportion of underfilled packets. c) \(\left(13, \frac{1}{\sqrt{2\pi e}}\right)\) and \(\left(15, \frac{1}{\sqrt{2\pi e}}\right)\)

- Locate the maximum of the normal density. - Relate the inflection points to \(\mu\pm\sigma\). - Recall the central one- and two-standard-deviation percentages. - Use symmetry to find a one-sided tail probability.

1. a) The density is symmetric and reaches its maximum at \(x=4\), so \(\mu=4\). 2. A normal density changes concavity at \(\mu-\sigma\) and \(\mu+\sigma\). The graph’s inflection points are at \(x=3\) and \(x=5\), so \(\sigma=1\). 3. b) 1. The interval \([3, 5]=[\mu-\sigma, \mu+\sigma]\), so its probability is approximately \(68.3\%\). 4. b) 2. The interval \([2, 6]=[\mu-2\sigma, \mu+2\sigma]\), so its probability is approximately \(95.4\%\). 5. b) 3. The area from \(\mu\) to \(\mu+\sigma\) is approximately \(68.3\%\div2=34.15\%\). Therefore, \(P(X\ge5)\approx50\%-34.15\%=15.85\%\).

a) \(\mu=4\) and \(\sigma=1\) b) 1. Approximately \(68.3\%\) 2. Approximately \(95.4\%\) 3. Approximately \(15.85\%\)

- The x-coordinate of the peak is the mean. - The inflection points lie one standard deviation from the mean. - Express \(480\) as the mean minus a multiple of the standard deviation. - Use symmetry to divide the probability outside the central interval between the two tails.

1. a) The peak occurs at \(x=500\), so \(\mu=500\,\text{mL}\). 2. The inflection points occur at \(x=490\) and \(x=510\), each \(10\,\text{mL}\) from the mean. Therefore, \(\sigma=10\,\text{mL}\). 3. b) The cutoff is \(480=500-2\cdot10=\mu-2\sigma\). 4. By the Empirical Rule, about \(95\%\) of values lie within two standard deviations of the mean. The remaining \(5\%\) is split equally between the tails, so \(P(X<480)\approx0.025\).

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

- Identify the mean and standard deviation. - Express each fill volume as a number of standard deviations from the mean. - Recall the empirical-rule percentages for one and two standard deviations. - Use symmetry to split the probability outside a central interval.

1. a) The interval \([495, 505]\) is \([\mu-\sigma, \mu+\sigma]\). By the empirical rule, its probability is approximately \(68.3\%\). 2. b) The cutoff \(490\) is \(\mu-2\sigma\). Approximately \(95.4\%\) of the distribution lies between \(\mu-2\sigma\) and \(\mu+2\sigma\). 3. The total probability outside that interval is approximately \(100\%-95.4\%=4.6\%\). By symmetry, the left tail contains half, or approximately \(2.3\%\). 4. Since \(2.3\%<2.5\%\), the claim is justified.

a) Approximately \(68.3\%\) b) \(P(X<490)\approx2.3\%<2.5\%\)

- The negative squared exponent is greatest when the squared expression is zero. - Use symmetry together with the Empirical Rule. - Identify each cutoff as a number of standard deviations from the mean. - Changing only the mean shifts a normal density horizontally.

1. a) The squared term \((x-755)^2\) is minimized at \(x=755\), making the exponent equal to \(0\). Since \(e^0=1\), this gives the maximum density. 2. The maximum is \(f(755)=\frac{1}{5\sqrt{2\pi}}\approx0.0798\). On the graph, this is the peak at \(x=755\), just below \(y=0.08\). 3. b) 1) The cutoff is \(750=\mu-\sigma\). By the Empirical Rule, about \(68\%\) lies within one standard deviation, leaving \(32\%\) outside. Symmetry gives about \(16\%\) below \(750\). 4. b) 2) The interval is \([\mu-2\sigma, \mu+2\sigma]\), so its probability is about \(95\%\). 5. c) Decreasing the mean by \(5\,\text{mL}\) shifts the entire graph \(5\) units left without changing its shape or height. 6. Under the new setting, \(745=750-5=\mu_{\text{new}}-\sigma\), so about \(16\%\) of bottles contain less than \(745\,\text{mL}\).

a) The maximum occurs at \(x=755\), and \(f(755)\approx0.0798\). b) 1) Approximately \(16\%\) b) 2) Approximately \(95\%\) c) The graph shifts \(5\) units 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 to split the probability outside the central interval. - Changing the mean affects location; changing the standard deviation affects spread. - Relate each graph change to the area left of the fixed cutoff.

1. a) The cutoff is \(1000=1005-5=\mu-\sigma\). 2. About \(68\%\) of values lie within one standard deviation, so about \(32\%\) lie outside. By symmetry, about \(16\%\) lie below \(\mu-\sigma\). Thus, the underfilled probability is approximately \(0.16\). 3. b) 1) Increasing \(\mu\) shifts the density graph to the right without changing its shape. The cutoff \(1000\) then lies farther into the left tail, so the area to its left decreases. 4. b) 2) Decreasing \(\sigma\) makes the graph narrower and taller around the same mean. The values are more concentrated near \(1005\), so the area below \(1000\) decreases.

a) Approximately \(0.16\), or \(16\%\) b) 1) Increasing the mean shifts the graph right and reduces the area below \(1000\). b) 2) Decreasing the standard deviation makes the graph narrower and taller, concentrating more values near the mean and reducing the area below \(1000\).

- The highest point of the curve identifies the mean. - Inflection points occur one standard deviation from the mean. - Probability is represented by area under the density curve. - Use symmetry about the vertical line through the mean.

1. a) The peak occurs at \(x=10\), so \(\mu=10\,\text{cm}\). The inflection points are approximately \(x=8\) and \(x=12\), so \(\sigma\approx2\,\text{cm}\). 2. b) 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. c) By symmetry, half of the total area lies at or above the mean, so \(P(X\ge10)=0.5\). 4. d) By symmetry, \(P(10\le X\le12)\approx\frac{0.68}{2}=0.34\). 5. Therefore, \(P(X>12)=P(X\ge10)-P(10\le X\le12)\approx0.50-0.34=0.16\).

a) \(\mu=10\,\text{cm}\), \(\sigma\approx2\,\text{cm}\) b) Approximately \(0.68\) c) \(0.50\) d) Approximately \(0.16\)

- The peak of a normal density identifies the mean. - The inflection points are one standard deviation from the mean. - Convert the cutoff to a standard score. - Use an upper-tail probability.

1. The peak occurs at \(x=1000\), so \(\mu=1000\,\text{mL}\). 2. The inflection points are at approximately \(990\) and \(1010\), each \(10\,\text{mL}\) from the mean. Therefore, \(\sigma=10\,\text{mL}\). 3. The standard score for \(1015\) is \(z=\frac{1015-1000}{10}=1.5\). 4. Thus, \(P(X>1015)=1-\Phi(1.5)\approx0.0668\).

\(\mu=1000\,\text{mL}\), \(\sigma=10\,\text{mL}\), and \(P(X>1015)\approx0.0668\), or \(6.68\%\).

- The peak gives one distribution parameter. - The horizontal distance from the peak to an inflection point gives the standard deviation. - Standardize both specification limits. - Subtract cumulative probabilities to find an interval probability.

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

\(\mu=40\,\text{mm}\), \(\sigma=2\,\text{mm}\), and approximately \(0.8664\), or \(86.64\%\), of the parts meet specifications.

- Recall the maximum-value formula for a normal density. - Identify the interval that lies one standard deviation on either side of the mean. - An integral of a density function represents probability over an interval. - Express zero grams as a standard score to judge how likely a negative value is.

1. A normal density reaches its maximum at the mean, and the maximum value is \(\frac{1}{\sigma\sqrt{2\pi}}\). Thus, \(\frac{1}{20\sqrt{2\pi}}\approx0.01995\,\text{g}^{-1}\approx0.020\,\text{g}^{-1}\). 2. The interval \([140, 180]\) is \([\mu-\sigma, \mu+\sigma]\). For a normal distribution, \(P(140\le X\le180)\approx0.6827\), or about \(68.3\%\). 3. The integral is the probability that an orange has mass between \(160\,\text{g}\) and \((160+a)\,\text{g}\). By symmetry, the area from \(\mu\) to \(\mu+\sigma\) is about \(0.341\), so \(a=20\,\text{g}\). 4. The criticism is theoretically correct because the normal distribution has no lower bound. However, zero is \(8\) standard deviations below the mean, and \(P(X<0)\approx6.22\times10^{-16}\). The probability is negligible, so the model is reasonable for these oranges.

a) \(\frac{1}{20\sqrt{2\pi}}\approx0.01995\,\text{g}^{-1}\approx0.020\,\text{g}^{-1}\) b) About \(68.3\%\) c) It is the probability that an orange has mass from \(160\,\text{g}\) to \(180\,\text{g}\), and \(a=20\,\text{g}\). d) Negative values are theoretically possible, but their probability is negligible, so the model is reasonable.

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

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

a) Approximately \(0.133\) b) Approximately \(0.16\), or \(16\%\) c) \(k\approx6\,\text{mL}\); about \(95\%\) of bottles contain between \(998\,\text{mL}\) and \(1010\,\text{mL}\). d) A mean above the labeled amount lowers the 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 \(2.5\%\)

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

1. Set \(P(\mu-d\le X\le\mu+d)=0.95\). 2. After standardizing, \(P\left(-\frac{d}{\sigma}\le Z\le\frac{d}{\sigma}\right)=0.95\). 3. Let \(z=\frac{d}{\sigma}\). For a central \(95\%\) interval, \(2\Phi(z)-1=0.95\), so \(\Phi(z)=0.975\). 4. A table or inverse normal function gives \(z\approx1.96\). Therefore, \(d=z\sigma\approx1.96\cdot0.05=0.098\,\text{mm}\).

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

- Express a symmetric interval probability using \(\Phi\). - Find the standard normal quantile associated with a cumulative probability of \(0.995\). - Consider how decreasing \(\sigma\) changes the concentration around the mean. - Follow the instruction to round downward.

1. a) The standard score is \(z=\frac{199.0-200.5}{0.8}=-1.875\). Thus, \(P(L<199.0)=\Phi(-1.875)\approx0.0304\), or \(3.04\%\). 2. b) The tolerance interval is symmetric with radius \(2\,\text{cm}\). We need \(2\Phi\left(\frac{2}{\sigma}\right)-1\ge0.99\). 3. Therefore, \(\Phi\left(\frac{2}{\sigma}\right)\ge0.995\). The \(0.995\) quantile is approximately \(2.5758\). 4. Thus, \(\sigma\le\frac{2}{2.5758}\approx0.7764\,\text{cm}\). 5. Rounding down to two decimal places gives \(0.77\,\text{cm}\), which safely satisfies the requirement.

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

- Use the critical value associated with a central \(99\%\) interval. - For a one-sided tail, split the probability outside a symmetric interval equally. - In part c), identify the standard-deviation boundary whose cumulative probability is about \(0.977\).

1. a) A central \(99\%\) interval uses \(z^*\approx2.58\). Therefore, \(1200\pm2.58\cdot150=1200\pm387\), giving \([813, 1587]\). 2. b) Since \(1650=1200+3\cdot150=\mu+3\sigma\), the total probability outside the central three-standard-deviation interval is approximately \(0.003\). The right tail is half of that, so \(P(X>1650)\approx0.0015\). 3. c) The cumulative probability at \(\mu+2\sigma\) is approximately \(0.5+0.954\div2=0.977\). Therefore, \(c=1200+2\cdot150=1500\).

a) \([813, 1587]\) b) \(P(X>1650)\approx0.0015\) c) \(c=1500\)

- 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. - Find 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. Solving gives \(\sigma=\frac{12}{1.96}\approx6.12\,\text{g}\).

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

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

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

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

- Standardize the cutoff in part 1. - Use a binomial model for the count in part 2. - Express a centered interval as \([\mu-c, \mu+c]\). - Account for both the required probability and the integer-endpoint condition.

1. 1) Standardizing gives \(z=\frac{247-250}{2}=-1.5\). Therefore, \(P(X<247)=\Phi(-1.5)\approx0.0668\). 2. 2) Let \(Y\) be the number of underweight packages. Then \(Y\sim\operatorname{Binomial}(50,p)\), where \(p\approx0.0668\). 3. Thus, \(P(Y\le1)=(1-p)^{50}+50p(1-p)^{49}\approx0.1443\). 4. 3) Let the interval be \([250-c, 250+c]\). We need \(2\Phi\left(\frac{c}{2}\right)-1\ge0.95\), so \(\frac{c}{2}\ge z_{0.975}\approx1.96\). 5. Therefore, \(c\ge3.92\). The smallest integer value is \(c=4\), giving \([246\,\text{g}, 254\,\text{g}]\).

1) Approximately \(0.0668\), or \(6.68\%\) 2) Approximately \(0.1443\), or \(14.43\%\) 3) \([246\,\text{g}, 254\,\text{g}]\)

- Standardize endpoints that are written as multiples of \(\sigma\) from \(\mu\). - Think about where a symmetric bell curve should be centered to place the most area over a fixed interval. - Use the symmetry of a normal distribution. - Decide what fraction of the distribution lies to the left of its mean.

1. a) Standardizing the endpoints gives \(z=-1\) and \(z=1\). Therefore, \(P(\mu-\sigma\le Y\le\mu+\sigma)=\Phi(1)-\Phi(-1)=2\Phi(1)-1\approx0.6827\). The parameters cancel during standardization, so this probability is the same for every normal distribution. 2. b) For fixed \(\sigma\), the normal density is symmetric about \(\mu\). A fixed-width interval captures the greatest area when it is centered at the density peak. The midpoint of \([10, 20]\) is \(\frac{10+20}{2}=15\), so the maximizing value is \(\mu=15\). 3. c) Symmetry gives \(P(Y\le\mu)=0.5\) for every positive value of \(\sigma\). Doubling \(\sigma\) does not change this probability.

a) Approximately \(0.6827\) b) \(\mu=15\) c) The probability remains \(0.5\).

- Compare each probability with familiar central and half-central normal areas. - Determine whether the interval begins or ends at the mean. - Locate the relevant region under a bell curve. - For a one-sided boundary, identify the corresponding standard score.

1. a) The probability \(0.4985\) is approximately half of the central three-standard-deviation probability, since \(0.997\div2=0.4985\). 2. The interval width is \(1200-900=300=3\sigma\). Under this approximation, the interval can run from \(\mu\) to \(\mu+3\sigma\), giving \(\mu=900\), or from \(\mu-3\sigma\) to \(\mu\), giving \(\mu=1200\). 3. b) A left-tail probability of \(0.025\) corresponds to \(z\approx-1.96\). Thus, \(551=\mu-1.96\cdot100=\mu-196\). 4. Therefore, \(\mu=747\).

a) \(\mu=900\) or \(\mu=1200\) b) \(\mu=747\)

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

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

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

- Distinguish between the number of responses and the sample proportion. - Convert a predicted interval for the count to an interval for the proportion by dividing by the sample size. - Use the critical value for the central \(99\%\) of a normal distribution. - Compare the observed sample proportion with the predicted interval.

1. a) Under the claim, \(X\sim\operatorname{Bin}(600,0.85)\). The mean count is \(\mu=np=600\cdot0.85=510\), and the standard deviation is \(\sigma=\sqrt{600\cdot0.85\cdot0.15}=\sqrt{76.5}\approx8.746\). 2. The large-count condition is satisfied because \(np=510\) and \(n(1-p)=90\). 3. A central \(99\%\) normal interval uses \(z^*\approx2.576\). The predicted count interval is \(510\pm2.576\cdot8.746\approx[487.47, 532.53]\). 4. Dividing by \(600\) gives the sample-proportion interval \([0.8125, 0.8875]\), or approximately \([81.2\%, 88.8\%]\). 5. b) The observed proportion is \(\hat p=\frac{486}{600}=0.81\). Since \(0.81\) is below the interval, the result is not consistent with the company’s claim at this prediction level.

a) Approximately \([0.8125, 0.8875]\), or \([81.2\%, 88.8\%]\) b) No. The observed proportion \(0.81\) lies below the interval.

- 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 \((\mu, 0.5)\).