Question

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.4 in​, and a standard deviation given by sigma equals 2.1 in.

Complete parts a and b.

a. If 1 woman is randomly​ selected, find the probability that her height is between 61.6 in and 62.6 in.

The probability is approximately ____ ​(Round to four decimal places as​ needed.)

Answer 1

Given,

= 62.4 , = 2.1

We convert this to standard normal as

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

P(61.6 < X < 62.6 ) = P( X < 62.6) - P( X < 61.6)

= P( Z < 62.6 - 62.4 / 2.1) - P( Z < 61.6 - 62.4 / 2.1)

= P( Z < 0.0952) - P (Z < -0.3810)

= P( Z < 0.0952 ) - ( 1 - P( Z < 0.3810) )

= 0.5379 - ( 1 - 0.6484 )

= 0.1863