Question

Carrie attained a score of 650 on the verbal section of her SAT. The mean grade was a 500 with a standard deviation of 100. What percentage of test takers scored below and above Carrie?

a. 93 and 7
b. 85 and 15
c. 10 and 90
d. more information is required to answer this question

Answer 1

Approximately 93% of test takers scored below Carrie on the verbal section of the SAT, while 7% scored above her.

Step-by-step explanation:

To find the percentage of test takers who scored below and above Carrie on the verbal section of the SAT, we need to calculate the z-score for Carrie's score and use the standard normal distribution table. The formula to calculate the z-score is:

z = (x - μ) / σ

where x is the score, μ is the mean, and σ is the standard deviation. In this case, x = 650, μ = 500, and σ = 100. Calculating the z-score gives us:

z = (650 - 500) / 100 = 1.5

Using the standard normal distribution table, we can find that approximately 93% of test takers scored below Carrie and 7% scored above her.