Question 1

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4.5.Suppose 81 golfers played the course today. Find the probability that the average score of the 81 golfersexceeded 76.
A) 0.0228
B) 0.1293
C) 0.4772
D) 0.3707

Accepted Answer

To find the probability that the average score of the 81 golfers exceeded 76, we use the formula for the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the standard deviation and $n$ is the sample size. Here, $\sigma = 4.5$ and $n = 81$, so $SE = \frac{4.5}{\sqrt{81}} = \frac{4.5}{9} = 0.5$. To find the Z-score for a mean of 76, we use $Z = \frac{\bar{x} - \mu}{SE}$, where $\bar{x}$ is the sample mean (76) and $\mu$ is the population mean (75). Thus, $Z = \frac{76 - 75}{0.5} = 2$. Looking up the Z-score of 2 in the standard normal distribution table, we find that the area to the left of Z is 0.9772. Therefore, the probability that the average score exceeds 76 is $1 - 0.9772 = 0.0228$.

Question 2

The sampling distribution of the sample mean is shown. If the sample size is n = 16, what is the standarddeviation of the population from which the sample was drawn? Round to the nearest thousandth whereappropriate. 
A) $\sigma = 0.24$
B) $\sigma = 0.96$
C) $\sigma = 0.015$
D) $\sigma = 0.004$

Accepted Answer

A

Question 3

$\mu = 54 , \sigma = 32 , n = 64$
A) $\mu \frac { \ } { x } = 54 , \sigma \frac {  } { x } = 4$
B) $\mu _{\bar { x }} = 54 , \sigma_{ \bar { x } }= 32$
C) $\mu _{\bar { x }} = 54 , \sigma  _{\bar { x }} = 0.5$ 
D) $\mu  _{\bar { x }}= 6.75 , \sigma  _{\bar { x }}= 4$

Accepted Answer

The correct choice is A because it correctly applies the formulas for the mean ($\mu_{\bar{x}}$) and standard deviation ($\sigma_{\bar{x}}$) of the sampling distribution of the sample mean. The mean of the sampling distribution ($\mu_{\bar{x}}$) is equal to the population mean ($\mu$), which is 54. The standard deviation of the sampling distribution ($\sigma_{\bar{x}}$) is equal to the population standard deviation ($\sigma$) divided by the square root of the sample size ($n$), which is $32 / \sqrt{64} = 4$.

Question 4

What are the values of $\mu  _{\bar { x }} \text { and } \sigma _{\bar { x }}$
 for the sampling distribution of the sample mean shown?

A) $\mu  _{\bar { x }} = 8 , \sigma  _{\bar { x }} = 0.04$

B) $\mu  _{\bar { x }} = 8 , \sigma  _{\bar { x }}
 = 0.12$
C) $\mu _{\bar { x }} = 0.04 , \sigma _{\bar { x }}= 8$

D) $\mu _{\bar { x }}= 8 , \sigma  _{\bar { x }} = 0.08$

Accepted Answer

The answer of What are the values of \(\mu ...

Question 5

What are the values of $\mu  _{\bar { x }} \text { and } \sigma  _{\bar { x }}$ 
 for the sampling distribution of the sample mean shown?

A) $\mu  _{\bar { x }} = 390 , \sigma  _{\bar { x }} = 40$

B) $\mu  _{\bar { x }} = 390 , \sigma  _{\bar { x }} = 120$

C) $\mu  _{\bar { x }}= 40 , \sigma  _{\bar { x }} = 390$

D) $\mu _{\bar { x }} = 390 , \sigma _{\bar { x }}= 80$

Accepted Answer

The answer of What are the values of \(\mu ...

Question 6

The standard error of the mean is given by
A) $\frac { \sigma } { \sqrt { n } }$
B) $\mu - \bar { x }$
C) $| \mu - \bar { x } |$
D) $\mu \pm \sigma$ 2 Describe the distribution of the sample mean: nonnormal population.

Accepted Answer

The standard error of the mean is calculated as the population standard deviation ($\sigma$) divided by the square root of the sample size ($n$), which is accurately represented by option A: $\frac { \sigma } { \sqrt { n } }$. This formula is used to estimate the variability of the sample mean around the population mean.

Question 7

According to the law of large numbers, as more observations are added to the sample, the difference betweenthe sample mean and the population mean
A) Tends to become smaller
B) Tends to become larger
C) Remains about the same
D) Is inversely affected by the data added

Accepted Answer

The law of large numbers states that as the size of the sample increases, the sample mean will get closer to the population mean, hence the difference between them tends to become smaller.

Question 8

A farmer was interested in determining how many grasshoppers were in his field. He knows that thedistribution of grasshoppers may not be normally distributed in his field due to growing conditions. As hedrives his tractor down each row he counts how many grasshoppers he sees flying away. After several rows hefigures the mean number of flights to be 57 with a standard deviation of 12. What is the probability of thefarmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor?
A) 0.0571
B) 0.4429
C) 0.9429
D) 0.5710

Accepted Answer

Using the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normal with mean 57 and standard deviation 12/sqrt(40). The Z-score for 60 is (60-57)/(12/sqrt(40)) = 1.58. Using the standard normal distribution table, the probability of Z being greater than 1.58 is approximately 0.0571.

Question 9

The amount of corn chips dispensed into a 16-ounce bag by the dispensing machine has been identified aspossessing a normal distribution with a mean of 16.5 ounces and a standard deviation of 0.2 ounce. Suppose100 bags of chips were randomly selected from this dispensing machine. Find the probability that the samplemean weight of these 100 bags exceeded 16.6 ounces.
A) approximately 0
B) 0.1915
C) 0.3085
D) 0.6915

Accepted Answer

The standard error of the mean is 0.2/sqrt(100) = 0.02. The z-score for 16.6 is (16.6-16.5)/0.02 = 5. A z-score of 5 corresponds to a probability of approximately 0, meaning it's extremely unlikely for the sample mean to exceed 16.6 ounces.

Question 10

$$\mu = 28 , \sigma = 8 , \mathrm { n } = 22$$
A)
$\mu _{\bar { x }}= 28 , \sigma   _{\bar { x }} = 1.706$  
B)
$\mu _{\bar { x }} = 28 , \sigma  _{\bar { x }}= 0.364$

C)
$\mu  _{\bar { x }} = 28 , \sigma  _{\bar { x }} = 8$

D) 
$\mu _{\bar { x }}= 16.166 , \sigma _{\bar { x }}= 1.706$

Provide an appropriate response.

Accepted Answer

The mean of the sampling distribution of the sample mean ($\mu_{\bar{x}}$) is equal to the population mean ($\mu$), which is 28. The standard deviation of the sampling distribution of the sample mean ($\sigma_{\bar{x}}$) is equal to the population standard deviation ($\sigma$) divided by the square root of the sample size ($n$), which calculates to $8 / \sqrt{22}$ or approximately 1.706.

Question 11

Furnace repair bills are normally distributed with a mean of 267 dollars and a standard deviation of 20 dollars.If 64 of these repair bills are randomly selected, find the probability that they have a mean cost between 267dollars and 269 dollars.
A) 0.2881
B) 0.7881
C) 0.2119
D) 0.5517

Accepted Answer

The answer of Furnace repair bills are normally distributed with...

Question 12

The amount of money collected by a snack bar at a large university has been recorded daily for the past fiveyears. Records indicate that the mean daily amount collected is $4000 and the standard deviation is $450. Thedistribution is skewed to the right due to several high volume days (including football game days). Supposethat 100 days were randomly selected from the five years and the average amount collected from those dayswas recorded. Which of the following describes the sampling distribution of the sample mean?
A) normally distributed with a mean of $4000 and a standard deviation of $45
B) normally distributed with a mean of $4000 and a standard deviation of $450
C) normally distributed with a mean of $400 and a standard deviation of $45
D) skewed to the right with a mean of $4000 and a standard deviation of $450

Accepted Answer

The sampling distribution of the sample mean will be approximately normally distributed due to the Central Limit Theorem, which applies because the sample size is large (n=100). The mean of the sampling distribution is equal to the population mean ($4000), and the standard deviation of the sampling distribution (standard error) is the population standard deviation ($450) divided by the square root of the sample size (√100), which equals $45.

Question 13

Assume that the heights of men are normally distributed with a mean of 70.7 inches and a standard deviationof 2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height greater than71.7 inches.
A) 0.0021
B) 0.0210
C) 0.9005
D) 0.9979

Accepted Answer

To find the probability that the mean height of 64 randomly selected men is greater than 71.7 inches, we use the formula for the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 2.8$ inches and $n = 64$, so $SE = \frac{2.8}{\sqrt{64}} = \frac{2.8}{8} = 0.35$ inches. We then calculate the z-score for a mean height of 71.7 inches: $z = \frac{71.7 - 70.7}{0.35} = \frac{1}{0.35} \approx 2.857$. Looking up this z-score in a standard normal distribution table or using a calculator gives a probability of about 0.9979 that a z-score is less than 2.857, so the probability of it being greater is $1 - 0.9979 = 0.0021$.

Question 14

One year, professional sports players salaries averaged $1.9 million with a standard deviation of $0.7 million.Suppose a sample of 100 major league players was taken. Find the approximate probability that the averagesalary of the 100 players exceeded $1.1 million.
A) approximately 1
B) approximately 0
C) 0.2357
D) 0.7357

Accepted Answer

Given the large sample size (n=100) and the central limit theorem, the sampling distribution of the mean will be approximately normally distributed. The probability that the average salary exceeds $1.1 million is very high, almost certain, because $1.1 million is significantly below the population mean of $1.9 million.

Question 15

The number of violent crimes committed in a day possesses a distribution with a mean of 2.2 crimes per dayand a standard deviation of 6 crimes per day. A random sample of 100 days was observed, and the meannumber of crimes for the sample was calculated. Describe the sampling distribution of the sample mean.
A) approximately normal with mean = 2.2 and standard deviation = 0.6
B) approximately normal with mean = 2.2 and standard deviation = 6
C) shape unknown with mean = 2.2 and standard deviation = 6
D) shape unknown with mean = 2.2 and standard deviation = 0.6

Accepted Answer

The sampling distribution of the sample mean will be approximately normal due to the Central Limit Theorem, with the same mean as the population mean (2.2) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (6/√100 = 0.6).

Question 16

Assume that blood pressure readings are normally distributed with a mean of 122 and a standard deviation of6.4. If 64 people are randomly selected, find the probability that their mean blood pressure will be less than 124.
A) 0.9938
B) 0.0062
C) 0.9998
D) 0.8615

Accepted Answer

To find the probability that the mean blood pressure of 64 randomly selected people is less than 124, we use the formula for the z-score of a sample mean: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, 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: $z = \frac{124 - 122}{6.4/\sqrt{64}} = \frac{2}{0.8} = 2.5$. Looking up a z-score of 2.5 in the standard normal distribution table gives a probability of approximately 0.9938, indicating the probability that their mean blood pressure will be less than 124.

Question 17

A random sample of size n is to be drawn from a population with μ= 600 and σ = 300. What size samplewould be necessary in order to ensure a standard error of 20?

Accepted Answer

The answer of A random sample of size n is...

Question 18

Suppose a population has a mean of 7 for some characteristic of interest and a standard deviation of 9.6. Asample is drawn from this population of size 64. What is the standard error of the mean?
A) 1.2
B) 0.7
C) 0.15
D) 3.3

Accepted Answer

The standard error of the mean is calculated as the standard deviation of the population divided by the square root of the sample size. Here, it is 9.6 / sqrt(64) = 9.6 / 8 = 1.2.

Question 19

Suppose a population has a mean weight of 180 pounds and a standard deviation of 25 pounds. A sample of100 items is drawn from this population. What is the standard error of the mean?
A) 2.5
B) 1.8
C) 18.0
D) 0.25

Accepted Answer

The standard error of the mean is calculated by dividing the population standard deviation by the square root of the sample size: SE = σ / √n. Here, σ = 25 pounds and n = 100. Thus, SE = 25 / √100 = 25 / 10 = 2.5.

Question 20

The sampling distribution of the sample mean is shown. If the sample size is n = 36, what is the standarddeviation of the population from which the sample was drawn? Round to the nearest thousandth whereappropriate. 
A) $\sigma = 300$
B) $\sigma = 1800$
C) $\sigma = 8.333$
D) $\sigma = 1.389$

Accepted Answer

The answer of The sampling distribution of the sample mean...

Quiz 8: Sampling Distributions

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4.5.Suppose 81 golfers played the course today. Find the probability that the average score of the 81 golfersexceeded 76.

The sampling distribution of the sample mean is shown. If the sample size is n = 16, what is the standarddeviation of the population from which the sample was drawn? Round to the nearest thousandth whereappropriate.

μ=54,σ=32,n=64\mu = 54 , \sigma = 32 , n = 64μ=54,σ=32,n=64

What are the values of μxˉ and σxˉ\mu _{\bar { x }} \text { and } \sigma _{\bar { x }}μxˉ​ and σxˉ​ for the sampling distribution of the sample mean shown?

What are the values of μxˉ and σxˉ\mu _{\bar { x }} \text { and } \sigma _{\bar { x }}μxˉ​ and σxˉ​ for the sampling distribution of the sample mean shown?

The standard error of the mean is given by

According to the law of large numbers, as more observations are added to the sample, the difference betweenthe sample mean and the population mean

μ=28,σ=8,n=22\mu = 28 , \sigma = 8 , \mathrm { n } = 22μ=28,σ=8,n=22

Furnace repair bills are normally distributed with a mean of 267 dollars and a standard deviation of 20 dollars.If 64 of these repair bills are randomly selected, find the probability that they have a mean cost between 267dollars and 269 dollars.

Assume that the heights of men are normally distributed with a mean of 70.7 inches and a standard deviationof 2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height greater than71.7 inches.

One year, professional sports players salaries averaged $1.9 million with a standard deviation of $0.7 million.Suppose a sample of 100 major league players was taken. Find the approximate probability that the averagesalary of the 100 players exceeded $1.1 million.

Assume that blood pressure readings are normally distributed with a mean of 122 and a standard deviation of6.4. If 64 people are randomly selected, find the probability that their mean blood pressure will be less than 124.

A random sample of size n is to be drawn from a population with μ= 600 and σ = 300. What size samplewould be necessary in order to ensure a standard error of 20?

Suppose a population has a mean of 7 for some characteristic of interest and a standard deviation of 9.6. Asample is drawn from this population of size 64. What is the standard error of the mean?

Suppose a population has a mean weight of 180 pounds and a standard deviation of 25 pounds. A sample of100 items is drawn from this population. What is the standard error of the mean?

The sampling distribution of the sample mean is shown. If the sample size is n = 36, what is the standarddeviation of the population from which the sample was drawn? Round to the nearest thousandth whereappropriate.

$\mu = 54 , \sigma = 32 , n = 64$

$\mu = 28 , \sigma = 8 , \mathrm { n } = 22$