Heights of women are normally distributed with a mean of 635

Heights of women are normally distributed with a mean of 63.5 inches and a standard deviation of 2.5 inches. Draw a sketch of the distribution of women\'s heights, labelling and

on your sketch. Use this information and the Empirical Rule to answer a) to d), shading the areas that represent the percentages/probabilities you have found.

Approximately what percentage of women is taller than 71 inches?  

What is the approximate probability that a randomly selected woman is between 56 and 58.5 inches tall?

What is the maximum possible height of a woman who falls within the shortest 2.5% of all women?

In a group of 1000 women, approximately how many would you expect to be between 58.5 and 66 inches tall?

Solution

Mean ( u ) =63.5
Standard Deviation ( sd )=2.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 71) = (71-63.5)/2.5
= 7.5/2.5 = 3
= P ( Z >3) From Standard Normal Table
= 0.0013                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 56) = (56-63.5)/2.5
= -7.5/2.5 = -3
= P ( Z <-3) From Standard Normal Table
= 0.00135
P(X < 58.5) = (58.5-63.5)/2.5
= -5/2.5 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(56 < X < 58.5) = 0.02275-0.00135 = 0.0214                  
c)
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 63.5/2.5 ) = 0.025
That is, ( x - 63.5/2.5 ) = -1.96
--> x = -1.96 * 2.5 + 63.5 = 58.6                  
d)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 58.5) = (58.5-63.5)/2.5/ Sqrt ( 1000 )
= -5/0.0791
= -63.2456
= P ( Z <-63.2456) From Standard Normal Table
= 0
P(X < 66) = (66-63.5)/2.5/ Sqrt ( 1000 )
= 2.5/0.0791 = 31.6228
= P ( Z <31.6228) From Standard Normal Table
= 1
P(58.5 < X < 66) = 1-0 = 1