Assume that womens heights are normally distributed with a m

Assume that women\'s heights are normally distributed with a mean given by mean = 62.1in, and a standard deviation given by variance of 3.1in,

A) If one woman is randomly selected find the probabllity that her height is less than 63in.    ________?

B) If 41 woman are randomly selected find the probablity that they have a mean height of less than 63in.

Solution

Normal Distribution
Mean ( u ) =62.1
Standard Deviation ( sd )=1.8
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 63) = (63-62.1)/1.8
= 0.9/1.8= 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.6915                  
b)
P(X < 63) = (63-62.1)/1.8/ Sqrt ( 41 )
= 0.9/0.2811= 3.2016
= P ( Z <3.2016) From Standard NOrmal Table
= 0.9993