Question

a consider the normal distribution of IQs with a mean of 100 and a standard deviation of 16. What

percentage of IQs are?

a. greater than 95?

b. less than 120

c. between 90 and 110

Answer 1

Population mean $\mu=100$

Standard deviation $\sigma=16$

(a) Greater than 95

$Z=\dfrac{X-\mu}{\sigma}$

$Z=\dfrac{95-100}{16}=-0.3125$

For values greater than 95

$P(1-Z)=1-0.37828$

$=0.62172$

Percentage

$=0.62172\times 100 =66.172\%$

(b) Less than 120

$Z=\dfrac{X-\mu}{\sigma}=\dfrac{120-100}{16}=1.25$

$P(Z=1.25)=0.89435$

Percentage

$=0.89435\times100=89.435\%$

(c) Between 90 and 110

$=\dfrac{90-100}{16}<Z<\dfrac{110-100}{16}$

$=P(-0.625<Z<0.625)$

$=P(Z<0.625)-P(Z<-0.625)$

$=0.73237-0.26763$

$=0.46474$

Percentage

$=0.46474\times100=46.474\%$