Question

Assume the heights in a female population are normally distributed with mean 65.7 inches and standard deviation 3.2 inches. Then, the probability that a typical female from this population is between 5 feet and 5 feet 7 inches tall is (to the nearest three decimals) ________.

Answer 1

0.620

Explanation:

We know that 1 feet = 12 inches, so, 5 feet is equivalent to 60 inches. Then, we are looking for the probability that a typical female from this population is between 60 inches and 67 inches. We know that

`\mu` = 65.7 inches and

`\sigma` = 3.2 inches

and the normal density function for this mean and standard deviation is

`(1)/(√(2\pi ) 3.2)exp[-((x-65.7)^(2))/(2(3.2)^(2)) ]`

The probability we are looking for is given by

`\int\limits^(67)_(60) {(1)/(√(2\pi ) 3.2)exp[-((x-65.7)^(2))/(2(3.2)^(2)) ] } \, dx =0.620`

You can use a computer to calculate this integral. You can use the following instruction in the R statistical programming language

pnorm(67, 65.7, 3.2)-pnorm(60, 65.7, 3.2)