Questions: On Jennifer's first math test she scored a 71. The mean and standard deviation for the class were 74 and 2.24, respectively. On her second math test, Jennifer scored a 75. The mean and standard deviation for the class were 77 and 1.63, respectively. On which test did Jennifer do better relative to the rest of the class? Explain your reasoning clearly in the space provided below. Show work on your scrap paper(worth 4 points), round all answers to 2 decimal places, if necessary.

Solution Steps

To determine Jennifer's performance on her first test, we calculate the z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \(X = 71\) (Jennifer's score)
• \(\mu = 74\) (mean score of the class)
• \(\sigma = 2.24\) (standard deviation of the class)

Substituting the values, we have:

\[ z = \frac{71 - 74}{2.24} = \frac{-3}{2.24} \approx -1.34 \]

Thus, the z-score for the first test is \(z \approx -1.34\).

Next, we calculate the z-score for Jennifer's second test using the same formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \(X = 75\) (Jennifer's score)
• \(\mu = 77\) (mean score of the class)
• \(\sigma = 1.63\) (standard deviation of the class)

Substituting the values, we have:

\[ z = \frac{75 - 77}{1.63} = \frac{-2}{1.63} \approx -1.23 \]

Thus, the z-score for the second test is \(z \approx -1.23\).

Now we compare the z-scores from both tests:

• Z-score for the first test: \(-1.34\)
• Z-score for the second test: \(-1.23\)

Since \(-1.23 > -1.34\), Jennifer performed better relative to her classmates on the second test.

Final Answer