Question

The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.6 years and a standard deviation of 1.9 years. Random samples of size 18 are drawn from the population and the mean of each sample is determined. Use the Central Limit Theorem to find the mean and standard error of the mean of the sampling distribution.

​(Round

to two decimal places as

needed​.)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 5.6

standard deviation = $\sigma$ = 1.9

n = 18

Central Limit Theorem

sampling distribution of mean

$\mu$$\bar x$ =  $\mu$ = 5.6

satnadard error

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 1.9 / $\sqrt$18 = 0.45