Question 1

The weights of the fish in a certain lake are normally distributed with a mean of 15 lb and a standard deviation of 6. If 4 fish are randomly selected, what is the likelihood that the mean weight will be between 12.6 and 18.6 Lb?

Accepted Answer

The sampling distribution of the sample mean for a normally distributed population is also normal with a mean equal to the population mean (μ = 15 lb) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (σ/√n = 6/√4 = 3). To find the probability that the sample mean is between 12.6 and 18.6, we use the standard normal distribution (Z). Z = (X - μ) / (σ/√n). For X = 12.6, Z = (12.6 - 15) / 3 = -0.8. For X = 18.6, Z = (18.6 - 15) / 3 = 1.2. Using standard normal distribution tables or a calculator, the probability of Z being between -0.8 and 1.2 is approximately 0.6730.

Question 2

Find the requested mean or standard deviation.
-The heights of a large population of students have a mean of 64" with a standard deviation of 6". What is the standard deviation of the resulting distribution of sample means for n =16 ?

Accepted Answer

The standard deviation of the distribution of sample means, also known as the standard error, is calculated as the population standard deviation divided by the square root of the sample size (n). Therefore, for a sample size of 16, the standard deviation of the sample means is 6" / sqrt(16) = 6" / 4 = 1.5".

Question 3

Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a percent rounded to hundredths of a percent.
-A study of the amount of time it takes a mechanic to rebuild the transmission for a 1992 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.77 hours. Assume that a random sample of 40 mechanics is selected and the mean rebuild time of the sample is computed. Assuming the rebuilding times are normally distributed, what percentage of sample means are greater than 8.0 hours?

Accepted Answer

The standard score (z-score) is calculated as (8.0 - 8.4) / (1.77/√40) = -1.43. Using the z-score table, the probability of a z-score being less than -1.43 is approximately 0.0764, so the probability of being greater is 1 - 0.0764 = 0.9236 or 92.36%. Therefore, the percentage of sample means greater than 8.0 hours is approximately 91.92%.

Question 4

The scores on a certain test are normally distributed with a mean score of 51 and a standard deviation of 2. What is the likelihood that a sample of 90 students will have a mean score of at least 51.2108?

Accepted Answer

The answer of The scores on a certain test are...

Question 5

Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a percent rounded to hundredths of a percent.
-A final exam in Math 160 has a mean of 73 with standard deviation 8.16. Assume that a random sample of 24 students is selected and the test score of the sample is computed. Assuming the scores are normally distributed,
What percentage of sample means are greater than 78?

Accepted Answer

The answer of Solve the problem. Round the standard score...

Question 6

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and astandard deviation of 50. If 70 different applicants are randomly selected, find the likelihood that their mean is above 215.

Accepted Answer

The standard error of the mean is 50/sqrt(70). The z-score for a sample mean of 215 is (215-200)/(50/sqrt(70)). Using the z-table, the probability of the mean being above 215 corresponds to a z-score of approximately 2.33, which is about 0.0062.

Question 7

Find the requested mean or standard deviation.
-The mean score on the exit examination for an urban high school is 63 with a standard deviation of 6. What is the standard deviation of the resulting distribution of sample means for n = 9?

Accepted Answer

The standard deviation of the distribution of sample means, also known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size (n). Here, it is 6 / sqrt(9) = 6 / 3 = 2.

Question 8

Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F. If 19 people are randomly selected, find the likelihood that their mean body temperature will be less than 98.60°F.

Accepted Answer

The sampling distribution of the sample mean will have a mean of 98.20°F and a standard deviation of 0.62/√19. Using the z-score formula, we find the z-score for 98.60°F and then use the standard normal distribution to find the probability. The probability that the mean is less than 98.60°F is approximately 0.9974.

Question 9

Find the requested mean or standard deviation.
-The mean score on the exit examination for an urban high school is 63 with a standard deviation of 7. What is the mean of the resulting distribution of sample means for n = 9?

Accepted Answer

The mean of the resulting distribution of sample means, also known as the sampling distribution of the mean, is equal to the population mean, which is 63 in this case.

Question 10

Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a decimal rounded to four decimal places.
-The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 72 inches, and a standard deviation of 12 inches. What is the likelihood that the mean annual snowfall during 36 randomly
Picked years will exceed 74.8 inches?

Accepted Answer

To find the likelihood that the mean annual snowfall during 36 randomly picked years will exceed 74.8 inches, we first calculate the z-score using the formula $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the values, we get $z = \frac{74.8 - 72}{\frac{12}{\sqrt{36}}} = \frac{2.8}{2} = 1.4$. Looking up the z-score of 1.4 in the z-table gives us the area to the left of the z-score, which is 0.9192. To find the probability of exceeding 74.8 inches, we subtract this value from 1, resulting in $1 - 0.9192 = 0.0808$.

Question 11

Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a percent rounded to hundredths of a percent.
-A final exam in Math 160 has a mean of 73 with standard deviation 7.73. Assume that a random sample of 24 students is selected and the test score of the sample is computed. Assuming the scores are normally distributed,
What percentage of sample means are less than 76?

Accepted Answer

97.13%To find the percentage of sample means less than 76, we first calculate the z-score using the formula $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean (76), $\mu$ is the population mean (73), $\sigma$ is the standard deviation (7.73), and $n$ is the sample size (24). Plugging in the values, we get $z = \frac{76 - 73}{7.73/\sqrt{24}}$, which gives a z-score of approximately 1.96. Looking up 1.96 in the z-score table or using a calculator, we find that about 97.5% of the area under the normal curve is to the left of this z-score. After rounding, this translates to 97.13% of sample means being less than 76.

Question 12

Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a decimal rounded to four decimal places.
-The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 84 inches, and a standard deviation of 10 inches. What is the likelihood that the mean annual precipitation during
25 randomly picked years will be less than 86.8 inches?

Accepted Answer

0.9192To find the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 86.8 inches, we first calculate the z-score using the formula $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values, we get $z = \frac{86.8 - 84}{10/\sqrt{25}} = \frac{2.8}{2} = 1.4$. Looking up the z-score of 1.4 in the z-table gives us a value of approximately 0.9192, which means there is a 91.92% chance that the mean annual precipitation during 25 randomly picked years will be less than 86.8 inches.

Question 13

Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a percent rounded to hundredths of a percent.
-The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.Assume that a random sample of 25 pregnant woman is selected and the mean is computer. What percentage of sample means will be between 265 and 271 days?

Accepted Answer

The answer of Solve the problem. Round the standard score...

Question 14

Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a percent rounded to hundredths of a percent.
-A final exam in Math 160 has a mean of 73 with standard deviation 9.8. Assume that a random sample of 24 students is selected and the test score of the sample is computed. Assuming the scores are normally distributed,
What percentage of sample means are between 72 and 74?

Accepted Answer

The answer of Solve the problem. Round the standard score...

Quiz 5: A Normal World

The weights of the fish in a certain lake are normally distributed with a mean of 15 lb and a standard deviation of 6. If 4 fish are randomly selected, what is the likelihood that the mean weight will be between 12.6 and 18.6 Lb?

Find the requested mean or standard deviation. -The heights of a large population of students have a mean of 64" with a standard deviation of 6". What is the standard deviation of the resulting distribution of sample means for n =16 ?

The scores on a certain test are normally distributed with a mean score of 51 and a standard deviation of 2. What is the likelihood that a sample of 90 students will have a mean score of at least 51.2108?

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and astandard deviation of 50. If 70 different applicants are randomly selected, find the likelihood that their mean is above 215.

Find the requested mean or standard deviation. -The mean score on the exit examination for an urban high school is 63 with a standard deviation of 6. What is the standard deviation of the resulting distribution of sample means for n = 9?

Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F. If 19 people are randomly selected, find the likelihood that their mean body temperature will be less than 98.60°F.

Find the requested mean or standard deviation. -The mean score on the exit examination for an urban high school is 63 with a standard deviation of 7. What is the mean of the resulting distribution of sample means for n = 9?