Question

Using z-scores for grades. At one university, the students are given z-scores at the end of each semester rather than the traditional GPAs. The mean and standard deviation of all students’ cumulative GPAs, on which the z-scores are based, are 2.7 and .5, respectively.

a. Translate each of the following z-scores to the corresponding GPA: z = 2.0, z = -1.0, z = .5, z = -2.5.

b. Students with z-scores below −1.6 are put on probation. What is the corresponding probationary GPA?

c. The president of the university wishes to graduate the top 16% of the students with cum laude honors and the top 2.5% with summa cum laude honors. Where (approximately) should the limits be set in terms of z-scores? In terms of GPAs? What assumption, if any, did you make about the distribution of the GPAs at the university?

Short answer

1. In terms of GPA, the scores will be 3.7, 2.2, 2.95 and 1.45, respectively.
2. 1.9
3. Z-score for the top 16% and the top 2.5% are 1 and 2, respectively, and GPA scores are 3.2 and 3.7, respectively.

Step-by-step solution

Calculation of the GPA scores

The translations from z-scores to GPA are shown below:

$$\begin{gathered} \text{x1 = Mean + z}\left(\text{Standard} \text{deviation}\right) \\ \text{= 2.7 + 2}\left(\text{0.5}\right) \\ \text{= 3.7} \\ \text{x2 = Mean + z}\left(\text{Standard} \text{deviation}\right) \\ \text{= 2.7 - 1}\left(\text{0.5}\right) \\ \text{= 2.2} \\ \text{x3 = Mean + z}\left(\text{Standard} \text{deviation}\right) \\ \text{= 2.7 + 0.5}\left(\text{0.5}\right) \\ \text{= 2.95} \\ \text{x4 = Mean + z}\left(\text{Standard} \text{deviation}\right) \\ \text{= 2.7 - 2.5}\left(\text{0.5}\right) \\ \text{= 1.45} \end{gathered}$$

Here x1 represents the corresponding GPA for z score of 2.

Here x2 represents the corresponding GPA for z score of -1.

Here x3 represents the corresponding GPA for z score of 0.5.

Here x4 represents the corresponding GPA for z score of 1.45.

Calculation of the probationary GPA score

The calculation of probationary GPA (x5) is shown below:

$$\begin{gathered} \text{x5 = Mean + z}\left(\text{Standard} \text{deviation}\right) \\ \text{= 2.7 - 1.6}\left(\text{0.5}\right) \\ \text{= 1.9} \end{gathered}$$

Determination of z-scores, GPA and assumption

For the top 16 percent, the z-score must be 1 for having 1 standard deviation above the mean and for the top 2.5 percent, the z-score must be 2 for having 2 standard deviations above the mean. The GPA (x6) for z-score of 1 and GPA (x7) for z-score of 2, respectively are calculated below:

$$\begin{gathered} \text{x6 = Mean + z}\left(\text{Standard} \text{deviation}\right) \\ \text{= 2.7 + 1}\left(\text{0.5}\right) \\ \text{= 3.2} \\ \text{x7 = Mean + z}\left(\text{Standard} \text{deviation}\right) \\ \text{= 2.7 + 2}\left(\text{0.5}\right) \\ \text{= 3.7} \end{gathered}$$

There have been no assumptions made for this for the purpose of calculation.