Suppose the salaries of university professors are approximately normally distributed with a mean of $65,000 and a standard deviation of $7,000. If a random sample of size 25 is taken and the mean is calculated, what is the probability that the mean value will be between $62,500 and $64,000

0.2009

Step-by-step explanation:

Mean (μ) = 65000

Standard deviation (σ) = 7000

n = 25

Let X be the random variable which is a measure of salaries of university professors

Z = (μ - x) / σ/√n

Pr (62500 ≤ x ≤ 64000) = ?

Pr ((65000 - 62500) / 7000/√25 ≤ z ≤ (65000 - 64000) / 7000/√25)

= Pr (2500 / 7000/5 ≤ z ≤ 1000 / 7000/5)

= Pr (2500 / 1400 ≤ z ≤ 1000/1400)

= Pr (1.79 ≤ z ≤ 0.714)

= Pr (0 ≤ z ≤ 1.79) - Pr (0 ≤ z ≤ 0.714)

From the normal distribution table we have

0.4633 - 0.2624

= 0.2009