Nina and Amy began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. Nina took a test in English and earned a 71.8, and Amy took a test in Social Studies and earned a 60.7. Use the fact that all the students' test grades in the English class had a mean of 71.7 and a standard deviation of 11.7, and all the students' test grades in Social Studies had a mean of 60.6 and a standard deviation of 10.5 to answer the following questions.
a) Calculate the z-score for Nina's test grade.
b) Calculate the z-score for Amy's test grade.
c) Which person did relatively better?
i. Nina
ii. Amy
iii. They did equally well.

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Question a:

Nina took a test in English and earned a 71.8. In the English class had a mean of 71.7 and a standard deviation of 11.7.

This means that [tex]X = 71.8, \mu = 71.7, \sigma = 11.7[/tex]

So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{71.8 - 71.7}{11.7}[/tex]

[tex]Z = 0.0085[/tex]

Question b:

Amy took a test in Social Studies and earned a 60.7. Students' test grades in Social Studies had a mean of 60.6 and a standard deviation of 10.5.

This means that [tex]X = 60.7, \mu = 60.6, \sigma = 10.5[/tex]

So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{60.7 - 60.6}{10.5}[/tex]

[tex]Z = 0.0095[/tex]

c) Which person did relatively better?

Amy had a higher z-score, so she did relatively better.