Assume that womens heights are normally distributed with a m

Assume that women\'s heights are normally distributed with a mean given by mu= 62.1 in. and a standard deviation given by alpha = 2. in. Complete parts a and b.m. If 1 woman is randomly selected, find the probability that her height is between 61.7 in and 62.7 in. The probability is approximately

Solution

Mean ( u ) =62.1
Standard Deviation ( sd )=2.1
Normal Distribution = Z= X- u / sd ~ N(0,1)                                  

To find P(a <= Z <=b) = F(b) - F(a)
P(X < 61.7) = (61.7-62.1)/2.1/ Sqrt ( 1 )
= -0.4/2.1
= -0.1905
= P ( Z <-0.1905) From Standard Normal Table
= 0.42447
P(X < 62.7) = (62.7-62.1)/2.1/ Sqrt ( 1 )
= 0.6/2.1 = 0.2857
= P ( Z <0.2857) From Standard Normal Table
= 0.61245
P(61.7 < X < 62.7) = 0.61245-0.42447 = 0.188