Answer to Question #125696 in Statistics and Probability for Brooke

Question

The mean salary offered to students who are graduating from Coastal State University this year is $24,245 , with a standard deviation of $3643. A random sample of 80 Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for these 80 students is $24,250 or less?

Answer 1

Given : mean= 24245, s= 3643,n= 80

To find : P(X$\leq$ 24250)

solution:

P($\overline{X}$ ≤ 24250)= P($\frac{\overline{X}-mean}{\frac{s}{\sqrt{n}}}\leq\frac{24250-24245}{\frac{3643}{\sqrt{80}}} )$

= P(Z$\leq$ 0.0123)

Using Z table ,$P(Z\leq 0.0123)$ = 0.5049.

Answer: The probability that the mean salary offer for these 80 students is $24,250 or less will be 0.5049.