Question

A new school initiative was to investigate what could be done to enhance the middle 40% of students at a school considered at risk. If the average score of the MCA exam was 77 with standard deviation of 8.3, what is the minimum and maximum score to be considered in the middle 40%?

Answer 1

Assuming that the data is normally distributed.

P(X < A) = P(Z < (A - mean)standard deviation)

Mean = 77

Standard deviation = 8.3

Let the minimum and maximum score to be considered in the middle 40% be M and N

P(X < M) = 0.5 - 0.4/2 = 0.3

P(Z < (M - 77)/8.3) = 0.3

Take value of Z corresponding to 0.3 from standard normal distribution table

(M - 77)/8.3 = -0.52

M = 72.68

P(X < N) = 0.5 + 0.4/2 = 0.7

P(Z < (N - 77)/8.3) = 0.7

Take value of Z corresponding to 0.7 from standard normal distribution table

(M - 77)/8.3 = 0.52

M = 81.32

Minimum and maximum score to be considered in the middle 40% is 72.68 and 81.32 respectively.