The distribution of the commute times for the employees at a large company has mean 22.4 minutes and standard deviation 6.8 minutes. A random sample of n employees will be selected and their commute times will be recorded. What is true about the sampling distribution of the sample mean as n increases from 2 to 10 ? The mean increases, and the variance increases. A The mean increases, and the variance decreases.
B The mean does not change, and the variance does not change.
C The mean does not change, and the variance increases.
D The mean does not change, and the variance decreases.

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 22.8, \sigma = 6.8[/tex]

By the Central Limit Theorem, the mean remains the mean no matter the size of the sample.

Now let's see what happens with the standard deviation(and the variance).

The variance is the standard deviation multiplied by itself, that is, elevated to the 2nd power. So as the standard deviation increases, so does the variance. As the standard deviation decreases, so does the variance.

n = 2

The standard deviation is

[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{6.8}{\sqrt{2}} = 4.8[/tex]

n = 10

[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{6.8}{\sqrt{10}} = 2.15[/tex]

So as the sample size increases, the standard deviation, and the variance, decreases.

So the correct answer is:

D The mean does not change, and the variance decreases.