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) What interval centered at a score of 100 contains 50% of the population?

c) 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$P(<X)=0.5-0.25$ to get the left value

$P(\frac{X-100}{15})=0.25$ X is less than 100

$P(Z<-y)=0.25$

$y=\frac{X-100}{15}$ from =NORM.S.INV(0.25) excel formula $y=-0.67448975$

thus $-0.67448975=\frac{X-100}{15}$

$X=(-0.67448975\times15)+100=89.88265$

The right limit is $(100-89.88265)+100=110.1173$

the interval {89.88,110.12} contains 50% of the population and is centered at a score of 100

c. P(>125)

$P(\frac{125-100}{15})=P(Z>)=1.667=0.0478$

$2500\times0.0478=119.5$

120 people out of 2500 will score above 125