Question

Suppose that scores on an aptitude test are normally distributed with a mean of 100 and a standard deviation of 4.6. Scores on a knowledge test are normally distributed with a mean of 70 and a standard deviation of 2.4. Felix scored 106 on the aptitude test and 75 on knowledge test. Which statement best describes Felix's scores on the two tests comparatively?
Answers:
Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the aptitude test, comparatively.


Felix's z-score on the aptitude test was −1.30 . His z-score on the knowledge test was −2.08 . Felix performed better on the aptitude test, comparatively.


Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.


Felix's z-score on the aptitude test was −1.30 . His z-score on the knowledge test was −2.08 . Felix performed better on the knowledge test, comparatively.

Answer 1

Answer:

The correct answer is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.

Step-by-step explanation:

1. Let's check all the information given to us to answer the question correctly:

Mean of the scores on the aptitude test = 100

Standard deviation of the aptitude test = 4.6

Felix's score on the aptitude test = 106

Mean of the scores on the knowledge test = 70

Standard deviation of the knowledge test = 2.4

Felix's score on the knowledge test = 75

2. Which statement best describes Felix's scores on the two tests comparatively?

Let's recall that z-score in a normal distribution, positive or negative, is the number of times of the standard deviation a certain element is from the mean. If the element is below the mean, then the z-score is negative and if it's above the mean, then the z-score is positive.

Therefore, a score of 106 on the aptitude test, will have the following z-score:

106 - 100 = 6 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

6/4.6 = 1.30

A score of 75 on the knowledge test, will have the following z-score:

75 - 70 = 5 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

5/2.4 = 2.08

The z-scores of Felix were 1.30 on the aptitude test and 2.08 on the knowledge test. He performed better on the aptitude test because 2.08 > 1.30, so the correct statement that best describes Felix's scores on the two tests comparatively is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.