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. What percent of candidates have a GPA above 3.9?

Answer 1

Answer:

Only 0.0668 of the candidates, or 6.68%, have a GPA above 3.9.

Step-by-step explanation:

The GPA, according to the question, is a random variable normally distributed. A normal distribution is determined by its parameters. These parameters are the population mean, $\\ \mu$, and the population standard deviation, $\\ \sigma$. The normal distribution for GPA has a $\\ \mu = 3.3$, and $\\ \sigma = 0.4$.

To find probabilities related to normally distributed data, we can use the standard normal distribution. To use it, we first need to find the corresponding standardized score or z-score for the value we want to obtain the probability. This value, in this case, is x = 3.9. The related z-score or standardized value is given by the formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

A z-score tells us the distance from the mean in standard deviations units. A negative value indicates that the value is below the mean. A positive value is, conversely, above the mean.

And we already have all the necessary data to use [1].

Thus

$\\ z = \frac{3.9 - 3.3}{0.4}$

$\\ z = \frac{0.6}{0.4}$

$\\ z = 1.5$

This value for z tells us that the "equivalent" raw score, x = 3.9, is above the mean, and it is at 1.5 standard deviations units from the population mean.

We can find the probabilities for standardized values in the cumulative standard normal table, available on the Internet or in Statistic books (we can also use statistical packages or even spreadsheets).

To use the cumulative standard normal table, we have the z-score as an entry. With this value, we find the z column on this table, and, since it is z = 1.5, we only need to select the first column (which has two decimal places, that is, .00). Then, the cumulative probability for P(z<1.50) = 0.9332.

However, we are asked for the percent of candidates that have a GPA above 3.9 (z-score = 1.50). This probability is the complement of P(z<1.50) or 1 - P(z<1.50). Mathematically

$\\ P(z3.9) = 1$

$\\ P(z1.50) = 1$ (standardized)

$\\ P(z>1.50) = 1 - P(z$

$\\ P(z>1.50) = 1 - 0.9332$

$\\ P(z>1.50) = 0.0668$

That is, only 0.0668 of the candidates, or 6.68%, have a GPA above 3.9.

We can see this probability in the graph below.