Question

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?

1. 0.7995
2. 0.9131
3. 0.2005
4. -0.2611

Answer 1

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