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? a. .1465 b. .0827 c. .0371 d. .2005

0.2005

Step-by-step explanation:

Mean, m = 65000

Standard deviation, σ= 7000

Sample size, n = 25

Let X = random variable of salary

Recall:

Z = (μ - x) /(σ/√n)

P(62500 ≤ x ≤ 64000) =?

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

P(2500 / 1400 ≤ z ≤ 1000/1400)

P = (1.79 ≤ z ≤ 0.714)

Using the normal distribution table or a Z probability calculator

0.4633 - 0.2624

= 0.2009

yes

yes yes yes

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.