Question

1. The IQs of all college freshmen last year were approximately normally distributed, with a mean of 115 and a standard deviation of 8. The middle 50% of the IQs are between what two IQs?

Answer 1

Solution:-

Given that,

mean = = 115

standard deviation = = 8

middle 50% is

P(-z Z z) = 0.50

P(Z z) - P(Z -z) = 0.50

2P(Z z) - 1 = 0.50

2P(Z z) = 1 + 0.50 = 1.50

P(Z z) = 1.50 / 2 = 0.75

P(Z 0.674) = 0.75

z = -0.674 , +0.674

Using z-score formula,

x = z * + = -0.674 *8 + 115 = 109.61

x = z * + = 0.674 *8 + 115 = 120.39

The IQs are between two values are = 109.61 and 120.39.