Question

Regina and Christina began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. Regina took a test in Art History and earned a 78, and Christina took a test in Math and earned a 64.4. Use the fact that all the students' test grades in the Art History class had a mean of 72.1 and a standard deviation of 11.9, and all the students' test grades in Math had a mean of 62.7 and a standard deviation of 8.5 to answer the following questions. a) Calculate the z-score for Regina's test grade. b) Calculate the z-score for Christina's test grad

Answer 1

Regina's z-score for her Art History test is approximately 0.50, indicating that her score is half a standard deviation above the class mean. Christina's z-score for her Math test is approximately 0.20, which is one-fifth a standard deviation above the class mean.

Step-by-step explanation:

To compare Regina's and Christina's test scores using a standardized measure, we calculate their z-scores. A z-score represents how many standard deviations an individual score is from the mean of the distribution.

a) Calculation of Regina's z-score

For Regina in Art History, the mean (μ) is 72.1 and the standard deviation (σ) is 11.9. Regina's score (X) is 78. The z-score is calculated using the formula:

Z = (X - μ) / σ

Plugging in the values:

Z = (78 - 72.1) / 11.9 ≈ 0.4958

Regina's z-score is approximately 0.50, meaning her score is about half a standard deviation above the class mean.

b) Calculation of Christina's z-score

For Christina in Math, the mean (μ) is 62.7 and the standard deviation (σ) is 8.5. Christina's score (X) is 64.4. The z-score is calculated using the formula:

Z = (X - μ) / σ

Plugging in the values:

Z = (64.4 - 62.7) / 8.5 ≈ 0.2

Christina's z-score is approximately 0.20, meaning her score is one-fifth a standard deviation above the class mean.