Womens heights are normally distributed with mean 631 in and

Women?s heights are normally distributed with mean 631 in and standard deviation of 2-5 in. A social organization for tall people has a requirement that women must be at least 69 in tat What percentage of women meet that requirement? Click to view page 1 of the table Click to view page 2 of the table The percentage of women that are taller than 69 in is %. (Round to two decimal places as needed.)

Solution

It is given that mu=63.1, sigma=2.5

X=69, convert the X value into z score:

Z=(X-mu)/sigma=(69-63.1)/2.5

=2.36

Now, P(x>69)=P(Z>2.36)

=0.9909

The required percentage is 99.09%