91Ó°ÊÓ

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.

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

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

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.