Question

A high-profile consulting company chooses its new entry-level employees from a pool of recent college graduates using a five-step interview process. Unfortunately, there are usually more candidates who complete the interview process than the number of new positions that are available. As a result, cumulative GPA is used as a tie-breaker. GPAs for the successful interviewees are Normally distributed, with a mean of 3.3 and a standard deviation of 0.4. Out of the163 people who made it through the interview process, only 121 can be hired. What cut-off GPA should the company use

Answer 1

Answer:

3.04

Step-by-step explanation:

Mean = 3.34

Standard deviation= 0.4

n = 163

X ~ N(3.3, 0.4)

The cut off gpa = j

P(x>j) = 121/163

P(x<=j) = 42/163 = 0.26

O.26 represents the 26th percentile

We solve further by using Microsoft Excel. We use the function NORMINV on excel to get GPA

X =. NORMINV(0.26, 3.3, 0.4)

X = 3.04 approximately

We conclude that the required gpa is 3.04