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.

•there must be variation for the particular trait within a population.

•the variation must be inheritable (that is, it must be capable of being passed on from the parents to their offspring).

•individuals with one version of the trait must produce more offspring than those with a different version of the trait.good

explanation: