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

(a) 0.5899

(b) 0.9166

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) = (1)/(√(2\pi)2.2)\exp{-((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/√(37)`. So, the probability density function is given by

`g(x) = (1)/(√(2\pi)(2.2)/(√(37)))\exp{-((x-62.5)^(2))/(2(2.2/√(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.