Question

Several intelligence tests follow a normal distribution with a mean of 100 and a standard deviation of 15.

a. Determine the percentage of the population that would obtain a score between 95 and 110.

b. For a population of 2,500, how many are expected to have a score above 125?

Answer 1

a. $P(95<X<110)$

$P(95<X<110)=P(\frac{95-100}{15}<Z<\frac{110-100}{15})$

$=P(Z<0.6667)-P(Z< -0.3333)$

$=0.7475-0.3694=0.3781$

Thus, 37.81% of the population would obtain scores between 95 and 110

b. $P(X>125)=1-P(X<125)$

$=1-P(Z<\frac{125-100}{15})=1-P(Z<\frac{5}{3})=1-0.9522=0.0478$

For a population of 2,500,

$2500\times0.0478=119.5$

Thus, 120 people out of 2500 will score above 125.