In: Statistics and Probability
The starting salaries of individuals with an MBA degree are normally distributed with a mean of $90,000 and a standard deviation of $20,000. Suppose we randomly select 16 of these individuals with an MBA degree. What is the probability that the average starting salary for these individuals is at least $85,800?
Solution :
Given that,
mean =
= 90,000
standard deviation =
= 20,000
n = 16
= 90,000
= (
/
n)
= (20,000 /
16 ) = 5000
P (
85,800 )
= 1 - P (
85,800 )
= 1 - P (
-
/
)
( 85,800 - 90,000 / 5000 )
= 1 - P( z
- 4200 / 5000 )
= 1 - P ( z
- 0.84 )
Using z table
= 1 - 0.2005
= 0.7995
Probability = 0.7995
Option 1 ) 0.7995 is correct