### EXAMPLE

Height of women follows normal distribution with mean 64.5 and standard deviation of 2.5 inches. Find

a) The probability that a woman is shorter than 70 in.

b) The probability that a woman is between 60 and 70 in tall.

c) What is the height 10% of women are shorter than, i.e. what is the 10th percentile of women heights?

SOLUTION. X= women height; X~N(64.5, 2.5).

a) P(X <70)=P(Z< (70-64.5)/2.5)=P(Z<2.2)=0.9861

b) P( 60 < X < 70) = P( (60-64.5)/2.5) < Z < (70-64.5)/2.5)=P(-1.8< Z < 2.2)= P( Z <-2.2) – P( Z < -1.8) = 0.9861 – 0.0359 = 0.9502.

c) 10th percentile of X =?

0.1=P( X< x) = P( Z< (x-65.5)/2.5), so -2.33=(x-65.5)/2.5; x=59.675.

10% of women are shorter than 59.675 in.