Answer to Question #325575 in Statistics and Probability for yvaine

Question

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

We have a normal distribution, $\mu=100, \sigma=16.$

Let's convert it to the standard normal distribution,

$z=\cfrac{x-\mu}{\sigma}.$

$a.\ \ z=\cfrac{95-100}{16}=-0.31;\\ P(X>95)=P(Z>-0.31)=\\ =1-P(Z<-0.31)=$

$=1-0.3783=0.6217=62.17\%$ ​(from z-table).

$b.\ \ z=\cfrac{120-100}{16}=1.25;\\ P(X<120)=P(Z<1.25)=0.8944=89.44\%.$

$c.\ \ z_1=\cfrac{90-100}{16}=-0.63;\\ z_2=\cfrac{110-100}{16}=0.63;\\ P(90<X<100)=P(-0.63<Z<0.63)=\\ =P(Z<0.63)-P(Z<-0.63)=\\ =0.7357-0.2643=0.4714=47.14\%.$