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

Answer:

0.620

Step-by-step 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

$\frac{1}{\sqrt{2\pi } 3.2}exp[-\frac{(x-65.7)^{2}}{2(3.2)^{2}} ]$

The probability we are looking for is given by

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

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)