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 163 people who made it through the interview process, only 121 can be hired. What cut-off GPA should the company use?

a. 3.00
b. 3.04
c. 3.08
d. 3.12

Answer 1

Answer:

Option B) 3.04

Step-by-step explanation:

We are given the following in the question:

Number of people interviewed,n = 163

Number of people hire, x = 121

Proportion of people hire =

$\dfrac{x}{n} = \dfrac{121}{163} = 0.742$

Mean, μ = 3.3

Standard Deviation, σ = 0.4

We are given that the distribution of GPA for the successful interviewees is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.742

$P( X > x) = P( z > \displaystyle\frac{x - 3.3}{0.4})=0.742$  

$= 1 -P( z \leq \displaystyle\frac{x - 3.3}{0.4})=0.742$  

$=P( z \leq \displaystyle\frac{x - 3.3}{0.4})=0.258$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\dfrac{x - 3.3}{0.4} = -0.650\\\\x = 3.04$

Thus, cut-off GPA should be 3.04.

Option B) 3.04