**Question: **The average annual salary of employees of a large chain of departmental
stores is stated to be 45,000 dollars with a standard deviation of
12,000 dollars. A random sample of 36 employees are selected and their
salaries noted. Assuming the distribution salaries is normally
distributed, find the probability that the average salary of the sample
will be between 40,000 and 44,000 dollars.

** Solution: **
The population distribution can be assumed to be normal. Hence
according to central limit theorem the sampling distribution of means is
approximately normal with $\mu _{\overline{x}}$ = 45,000 and standard
error $\sigma _{\overline{x}}$ =

$\frac{12000}{\sqrt{36}}$ = 2000

The z score the distribution is calculated using the formula

z =

$\frac{x-\mu _{\overline{x}}}{\sigma \overline{x}}$The z scores for x = 40,000 and x = 44,000 are

z

_{1} =

$\frac{40000-45000}{2000}$ = -2.5 and z

_{2} =

$\frac{44000-45000}{2000}$ = -0.5

Hence P(40,000 ≤ x ≤ 44,000) = P(-2.5 ≤ z ≤ -0.5) = 0.3023.

The area under the normal curve representing the probability is shown below.