Question

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.5 in​,and a standard deviation given by sigma equals 2.2 in.​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63in.​(b) If 37women are randomly​ selected, find the probability that they have a mean height less than 63in.​(​a)The probability is approximately nothing.​(Round to four decimal places as​ needed.)​(b) The probability is approximately nothing.​(Round to four decimal places as​ needed.)

Answer 1

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with  

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by  

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$, then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

   (in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$. So, the probability density function is given by

$g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}$, and

$P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx$ = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.