Answer:
O A. No, 10.6% of the class scored above her test score
Step-by-step 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.