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 1.9 in. ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in. ​(b) If 32 women are randomly​ selected, find the probability that they have a mean height less than 63 in. ​(​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

Given,

= 62.5 , = 1.9

We convert this to standard normal as

P(X < x) = P( Z < x - / )

a)

P( X < 63) = P( Z < 63 - 62.5 / 1.9)

= P( Z < 0.2632)

= 0.6038 (Probability calculated from Z table)

b) Using central limit theorem,

P( < x) = P( Z < x - / ( / sqrt(n) ))

So,

P( < 63) = P( Z < 63 - 62.5 / (1.9 / sqrt(32) ) )

= P( Z < 1.4886)

= 0.9317 (Probability calculated from Z table)