Question

suppose jack and diane are each attempting to use a simulation to describe the sampling distribution from a population that is skewed right with mean 60 and standard deviation 5. jack obtains 1000 random samples of size n = 3 from the population, finds the mean of the means, and determines the standard deviation of the means. diane does the same simulation, but obtains 1000 random samples of size n = 30 from the population. complete parts (a) through (c).

(a) which of the following statements would be true regarding the sampling distributions of the sample mean?

a. jacks distribution is expected to be skewed right, but less skewed than the original distribution. dianes distribution is expected to be approximately normal.

b. dianes distribution is expected to be skewed right, but less skewed than the original distribution. jacks distribution is expected to be approximately normal.

c. jacks distribution and dianes distribution are expected to be approximately normal. however, jacks will have the smaller standard deviation of the sample mean.

d. jacks distribution and dianes distribution are expected to be approximately normal. however, jacks will have the greater standard deviation of the sample mean.

(b) what do you expect the mean of jacks distribution to be? what do you expect the mean of dianes distribution to be?

jacks distribution is expected to have a mean of 60. dianes distribution is expected to have a mean of 60.

(type integers or decimals rounded to two decimal places as needed.)

(c) what do you expect the standard deviation of jacks distribution to be? what do you expect the standard deviation of dianes distribution to be?

jacks distribution is expected to have a standard deviation of. dianes distribution is expected to have a standard deviation of.

(type integers or decimals rounded to two decimal places as needed.)

Explanation:

Step1: Recall the formula for standard deviation of the sample mean

The formula for the standard deviation of the sample mean (\(\sigma_{\bar{x}}\)) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

Step2: Calculate for Jack's distribution

Given \(\sigma = 5\) and \(n = 3\) for Jack.

$$ \sigma_{\bar{x}}=\frac{5}{\sqrt{3}}\approx\frac{5}{1.732}\approx 2.89 $$

Step3: Calculate for Diane's distribution

Given \(\sigma = 5\) and \(n = 30\) for Diane.

$$ \sigma_{\bar{x}}=\frac{5}{\sqrt{30}}\approx\frac{5}{5.477}\approx 0.91 $$

Answer:

Jack's distribution is expected to have a standard deviation of \(2.89\). Diane's distribution is expected to have a standard deviation of \(0.91\).