Mr. Johnson's class is normally distributed with a class average of 75% and a standard deviation of 6. What are the quartiles of Mr. Johnson's class?

Answer 1

The median = mean = 2nd quartile = 75

3rd quartile:

Prob ( Z < z) = 0.75

z= 0.6748975

z = (x - mean)/stddev

z * stddev + mean = x

0.6748975 * 6 + 75 = 79.049385

1st quartile

Prob ( Z < z) = 0.25

z= -0.6748975 per normal table and property of bell curve

x = 70.9530615