Question 1

Solve the problem.
-Suppose that you wish to find $\mathrm { P } ( - 2 < \mathrm { x } < 2 )$ for a continuous uniform distribution having a minimum of $- 3$ and a maximum of 3 . If you incorrectly assume that the distribution is normal instead of uniform, will your answer be too big, too small, or will you still obtain the correct answer? Explain your thinking.

Accepted Answer

The answer of Solve the problem.
-Suppose that you wish to...

Question 2

The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. What percentage of trainees earn less than $900 a month?

Accepted Answer

To find the percentage of trainees earning less than $900, we calculate the z-score: $z = \frac{X - \mu}{\sigma} = \frac{900 - 1100}{150} = -\frac{200}{150} = -1.33$. Looking up -1.33 on the standard normal distribution table gives us approximately 9.18%.

Question 3

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\mu = 60.0$ and the standard deviation is $\sigma = 4.0$. Find the probability that $X$ is less than $53.0$.

Accepted Answer

The answer of Assume that X has a normal distribution,...

Question 4

Solve the problem.
-In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. Find P45, which is the consumption level separating the bottom 45% from the top 55%.

Accepted Answer

P45 corresponds to the 45th percentile of the distribution. To find this, we use the Z-score formula for a percentile in a normal distribution, which is Z = (X - μ) / σ, where X is the value we're looking for, μ is the mean, and σ is the standard deviation. However, since we're given the percentile (45th), we first find the Z-score corresponding to the 45th percentile from a Z-table or using a calculator. The Z-score for the 45th percentile is approximately -0.125. Plugging this into the formula and solving for X gives us X = Zσ + μ. Substituting the given values (μ = 1050, σ = 218, Z ≈ -0.125) results in X ≈ 1021.7, which corresponds to option B.

Question 5

Solve the problem.
-The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.1 and a standard deviation of 40.7. All units are in mg/100 mL. Find the two levels that separate the top 9% and the bottom 9%.

Accepted Answer

The answer of Solve the problem.
-The serum cholesterol levels for...

Question 6

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$. Find the probability that $X$ is greater than $15.2$.

Accepted Answer

Since $X$ has a normal distribution with a mean of $15.2$, the probability of $X$ being greater than the mean (15.2) is 0.5 or 50%. This is because, in a normal distribution, half of the values lie above the mean and half lie below the mean.

Question 7

Solve the problem.
-Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F. Find the temperature that separates the top 7% from the bottom 93%.

Accepted Answer

The answer of Solve the problem.
-Human body temperatures are normally...

Question 8

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\boldsymbol { \mu } = 137.0$ and the standard deviation is $\boldsymbol { \sigma } = 5.3$. Find the probability that $X$ is between $134.4$ and 140.1.

Accepted Answer

The probability that X is between 134.4 and 140.1 is the area under the normal distribution curve between those two values. Using a standard normal distribution table, we find that the area between z = -0.47 and z = 0.58 is 0.4069.

Question 9

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$. Find the probability that $X$ is between $14.3$ and 16.1.

Accepted Answer

The probability that $X$ is between $14.3$ and $16.1$ is found by calculating the z-scores for these values and using the standard normal distribution table. The z-scores are -1 and 1, respectively, and the probability between these z-scores is approximately 0.6826.

Question 10

The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What percentage of bolts will have a diameter greater than 0.32 inches?

Accepted Answer

To find the percentage of bolts with a diameter greater than 0.32 inches, we calculate the z-score for 0.32 inches using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation. Substituting the given values, we get $z = \frac{(0.32 - 0.30)}{0.01} = 2$. Looking up a z-score of 2 in a standard normal distribution table or using a calculator, we find that the area to the left of z=2 is approximately 0.9772, meaning 97.72% of the distribution is below 0.32 inches. Therefore, the percentage of bolts with a diameter greater than 0.32 inches is 100% - 97.72% = 2.28%.

Question 11

Solve the problem.
-The weights of certain machine components are normally distributed with a mean of 8.05 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. Theses weights could serve as limits used to identify which components should be rejected.

Accepted Answer

To find the weights that separate the top 3% and bottom 3%, we use the Z-score formula and the standard normal distribution. For the bottom 3%, we look for the Z-score corresponding to a cumulative probability of 0.03, which is approximately -1.88. For the top 3%, we look for the Z-score corresponding to a cumulative probability of 0.97 (since 1 - 0.03 = 0.97), which is approximately 1.88. Using the Z-score formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation, we solve for $X$ to find the weights.For the bottom 3%:$$ -1.88 = \frac{X - 8.05}{0.09} $$$$ X = -1.88 \times 0.09 + 8.05 $$$$ X \approx 7.88 \, g $$For the top 3%:$$ 1.88 = \frac{X - 8.05}{0.09} $$$$ X = 1.88 \times 0.09 + 8.05 $$$$ X \approx 8.22 \, g $$Therefore, the weights that separate the top 3% and bottom 3% are approximately 7.88 g and 8.22 g, which corresponds to choice B.

Question 12

Solve the problem.
-Scores on an English test are normally distributed with a mean of 33.8 and a standard deviation of 8.5. Find the score that separates the top 59% from the bottom 41%

Accepted Answer

To find the score that separates the top 59% from the bottom 41%, we need to find the z-score that corresponds to the 41st percentile. Using a standard normal distribution table or calculator, the z-score for 0.41 is approximately -0.23. Using the formula: score = mean + (z-score * standard deviation), we get: 33.8 + (-0.23 * 8.5) ≈ 31.8.

Question 13

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\mu = 22.0$ and the standard deviation is $\sigma = 2.4$. Find the probability that $\mathrm { X }$ is between $19.7$ and $25.3$.

Accepted Answer

To find the probability that $X$ is between 19.7 and 25.3, we use the standard normal distribution by converting the given values to Z-scores using the formula $Z = \frac{X - \mu}{\sigma}$. For $X = 19.7$, $Z = \frac{19.7 - 22.0}{2.4} = -0.9583$, and for $X = 25.3$, $Z = \frac{25.3 - 22.0}{2.4} = 1.375$. Looking these Z-scores up in a standard normal distribution table or using a calculator gives us the probability of $X$ being between these two values, which is approximately 0.7477.

Question 14

Solve the problem.
-The amount of rainfall in January in a certain city is normally distributed with a mean of 4.6 inches and a standard deviation of 0.3 inches. Find the value of the quartile $\mathrm { Q } _ { 1 }$

Accepted Answer

The first quartile $Q_1$ is the 25th percentile of a normal distribution. Using the Z-table, the Z-score for the 25th percentile is approximately -0.675. Using the formula $Q_1 = \mu + Z \cdot \sigma$, where $\mu = 4.6$ and $\sigma = 0.3$, we get $Q_1 = 4.6 + (-0.675) \cdot 0.3 = 4.4$.

Question 15

Solve the problem.
-Scores on a test are normally distributed with a mean of $65.3$ and a standard deviation of $10.3$. Find P81, which separates the bottom $81 \%$ from the top $19 \%$.

Accepted Answer

To find P81, which is the score that separates the bottom 81% from the top 19%, we use the Z-score formula and the normal distribution properties. The Z-score corresponding to 81% in the standard normal distribution table is approximately 0.88. We then use the formula for converting a Z-score to an actual score in a distribution: $X = \mu + Z\sigma$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $Z$ is the Z-score. Plugging in the values, we get $X = 65.3 + 0.88 \times 10.3 \approx 74.4$.

Question 16

Solve the problem.
-Suppose that replacement times for washing machines are normally distributed with a mean of 9.3 years and a standard deviation of 2 years. Find the replacement time that separates the top 18% from the bottom 82%.

Accepted Answer

To find the replacement time that separates the top 18% from the bottom 82%, we look for the z-score corresponding to the 82nd percentile in a standard normal distribution. Using a z-table or calculator, the z-score for the 82nd percentile is approximately 0.915. We then use the formula for converting a z-score to an original score: X = μ + zσ, where μ is the mean, σ is the standard deviation, and z is the z-score. Plugging in the values: X = 9.3 + 0.915(2) = 9.3 + 1.83 = 11.13, which rounds to 11.1 years.

Question 17

Solve the problem.
-Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the value of the quartile Q $Q _ { 3 }$

Accepted Answer

The third quartile (Q3) is the 75th percentile of a normal distribution. Using the Z-table, the Z-score for the 75th percentile is approximately 0.675. Using the formula: Q3 = mean + (Z * standard deviation), we get Q3 = 63.6 + (0.675 * 2.5) = 65.2875, which rounds to 65.3 inches.

Question 18

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$. Find the probability that $\mathrm { X }$ is greater than $16.1$.

Accepted Answer

To find the probability that X is greater than 16.1, we calculate the z-score: $ z = \frac{16.1 - 15.2}{0.9} = 1 $. The probability that Z is greater than 1 is 0.1587.

Question 19

Solve the problem.
-A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50 . Find $\mathrm { P } _ { 6 }$, the score which separates the lower $60 \%$ from the top $40 \%$.

Accepted Answer

The answer of Solve the problem.
-A bank's loan officer rates...

Question 20

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$. Find the probability that $X$ is greater than 17.

Accepted Answer

To find the probability that $X$ is greater than 17, we first calculate the z-score using the formula $z = \frac{X - \mu}{\sigma}$, where $X = 17$, $\mu = 15.2$, and $\sigma = 0.9$. This gives us $z = \frac{17 - 15.2}{0.9} = 2$. Looking up the z-score of 2 in the standard normal distribution table, we find the area to the left of $z = 2$ is approximately 0.9772. Since we want the area to the right (the probability that $X$ is greater than 17), we subtract this from 1, giving $1 - 0.9772 = 0.0228$.

Quiz 6: Normal Probability Distributions

The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. What percentage of trainees earn less than $900 a month?

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 60.0$ and the standard deviation is $\sigma = 4.0$ . Find the probability that $X$ is less than $53.0$ .

Solve the problem. -In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. Find P45, which is the consumption level separating the bottom 45% from the top 55%.

Solve the problem. -The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.1 and a standard deviation of 40.7. All units are in mg/100 mL. Find the two levels that separate the top 9% and the bottom 9%.

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $X$ is greater than $15.2$ .

Solve the problem. -Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F. Find the temperature that separates the top 7% from the bottom 93%.

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\boldsymbol { \mu } = 137.0$ and the standard deviation is $\boldsymbol { \sigma } = 5.3$ . Find the probability that $X$ is between $134.4$ and 140.1.

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $X$ is between $14.3$ and 16.1.

The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What percentage of bolts will have a diameter greater than 0.32 inches?

Solve the problem. -The weights of certain machine components are normally distributed with a mean of 8.05 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. Theses weights could serve as limits used to identify which components should be rejected.

Solve the problem. -Scores on an English test are normally distributed with a mean of 33.8 and a standard deviation of 8.5. Find the score that separates the top 59% from the bottom 41%

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 22.0$ and the standard deviation is $\sigma = 2.4$ . Find the probability that $\mathrm { X }$ is between $19.7$ and $25.3$ .

Solve the problem. -The amount of rainfall in January in a certain city is normally distributed with a mean of 4.6 inches and a standard deviation of 0.3 inches. Find the value of the quartile $\mathrm { Q } _ { 1 }$

Solve the problem. -Scores on a test are normally distributed with a mean of $65.3$ and a standard deviation of $10.3$ . Find P81, which separates the bottom $81 \%$ from the top $19 \%$ .

Solve the problem. -Suppose that replacement times for washing machines are normally distributed with a mean of 9.3 years and a standard deviation of 2 years. Find the replacement time that separates the top 18% from the bottom 82%.

Solve the problem. -Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the value of the quartile Q $Q _ { 3 }$

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $\mathrm { X }$ is greater than $16.1$ .

Solve the problem. -A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50 . Find $\mathrm { P } _ { 6 }$ , the score which separates the lower $60 \%$ from the top $40 \%$ .

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $X$ is greater than 17.

Quiz 6: Normal Probability Distributions

The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. What percentage of trainees earn less than $900 a month?

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=60.0\mu = 60.0μ=60.0 and the standard deviation is σ=4.0\sigma = 4.0σ=4.0 . Find the probability that XXX is less than 53.053.053.0 .

Solve the problem. -In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. Find P45, which is the consumption level separating the bottom 45% from the top 55%.

Solve the problem. -The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.1 and a standard deviation of 40.7. All units are in mg/100 mL. Find the two levels that separate the top 9% and the bottom 9%.

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=15.2\mu = 15.2μ=15.2 and the standard deviation is σ=0.9\sigma = 0.9σ=0.9 . Find the probability that XXX is greater than 15.215.215.2 .

Solve the problem. -Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F. Find the temperature that separates the top 7% from the bottom 93%.

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=137.0\boldsymbol { \mu } = 137.0μ=137.0 and the standard deviation is σ=5.3\boldsymbol { \sigma } = 5.3σ=5.3 . Find the probability that XXX is between 134.4134.4134.4 and 140.1.

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=15.2\mu = 15.2μ=15.2 and the standard deviation is σ=0.9\sigma = 0.9σ=0.9 . Find the probability that XXX is between 14.314.314.3 and 16.1.

The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What percentage of bolts will have a diameter greater than 0.32 inches?

Solve the problem. -Scores on an English test are normally distributed with a mean of 33.8 and a standard deviation of 8.5. Find the score that separates the top 59% from the bottom 41%

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=22.0\mu = 22.0μ=22.0 and the standard deviation is σ=2.4\sigma = 2.4σ=2.4 . Find the probability that X\mathrm { X }X is between 19.719.719.7 and 25.325.325.3 .

Solve the problem. -The amount of rainfall in January in a certain city is normally distributed with a mean of 4.6 inches and a standard deviation of 0.3 inches. Find the value of the quartile Q1\mathrm { Q } _ { 1 }Q1​

Solve the problem. -Scores on a test are normally distributed with a mean of 65.365.365.3 and a standard deviation of 10.310.310.3 . Find P81, which separates the bottom 81%81 \%81% from the top 19%19 \%19% .

Solve the problem. -Suppose that replacement times for washing machines are normally distributed with a mean of 9.3 years and a standard deviation of 2 years. Find the replacement time that separates the top 18% from the bottom 82%.

Solve the problem. -Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the value of the quartile Q Q3Q _ { 3 }Q3​

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=15.2\mu = 15.2μ=15.2 and the standard deviation is σ=0.9\sigma = 0.9σ=0.9 . Find the probability that X\mathrm { X }X is greater than 16.116.116.1 .

Solve the problem. -A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50 . Find P6\mathrm { P } _ { 6 }P6​ , the score which separates the lower 60%60 \%60% from the top 40%40 \%40% .

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=15.2\mu = 15.2μ=15.2 and the standard deviation is σ=0.9\sigma = 0.9σ=0.9 . Find the probability that XXX is greater than 17.

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 60.0$ and the standard deviation is $\sigma = 4.0$ . Find the probability that $X$ is less than $53.0$ .

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $X$ is greater than $15.2$ .

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\boldsymbol { \mu } = 137.0$ and the standard deviation is $\boldsymbol { \sigma } = 5.3$ . Find the probability that $X$ is between $134.4$ and 140.1.

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $X$ is between $14.3$ and 16.1.

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 22.0$ and the standard deviation is $\sigma = 2.4$ . Find the probability that $\mathrm { X }$ is between $19.7$ and $25.3$ .

Solve the problem. -The amount of rainfall in January in a certain city is normally distributed with a mean of 4.6 inches and a standard deviation of 0.3 inches. Find the value of the quartile $\mathrm { Q } _ { 1 }$

Solve the problem. -Scores on a test are normally distributed with a mean of $65.3$ and a standard deviation of $10.3$ . Find P81, which separates the bottom $81 \%$ from the top $19 \%$ .

Solve the problem. -Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the value of the quartile Q $Q _ { 3 }$

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $\mathrm { X }$ is greater than $16.1$ .

Solve the problem. -A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50 . Find $\mathrm { P } _ { 6 }$ , the score which separates the lower $60 \%$ from the top $40 \%$ .

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find the probability that $X$ is greater than 17.