A distribution of class scores on a test is approximately normal with a mean of 75 and a standard deviation of 8. Edna scores an 85%. Does her score place her in the top 10% of
this class?
O A. No, 10.6% of the class scored above her test score
B. Yes, 10.6% of the class scored above her test score
C. No, she scored better than 10.6% of her classmates
D. Yes, less than 10% of the class scored better than her test score

O A. No, 10.6% of the class scored above her test score

Explanation:

Using the standard normal distribution (Z score) to solve this problem. The formula is as follows:

Z = x - μ/σ

Where;

σ = standard deviation

μ = mean score

x = Edna's score

According to the information given in this question:

σ = 8

μ = 75%

x = 85%

Z = 85 - 75/8

Z = 10/8

Z = 1.25

Using the table of normal distribution probabilities, we check for Z score of 1.2 under 5 = 0.8944

This means that Edna's percentage score is 89.44%. Hence, (100 - 89.44) = 10.56% is above Edna's score in the class

Approximately, 10.6% of the class scored above Edna's test score, meaning that she is not among the top 10% of her class.