In a survey of women in a certain country 2029 the mean heig

In a survey of women in a certain country (20-29), the mean height was 63.4 inches with a standard deviation of 2.68 inches.? Answer the following questions about the specified normal distribution. A.) what height represents the 98th percentile? B.) what height represents the first quartile

Solution

Normal Distribution
Mean ( u ) =63.4
Standard Deviation ( sd )=2.68
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P ( Z < x ) = 0.98
Value of z to the cumulative probability of 0.98 from normal table is 2.054
P( x-u/s.d < x - 63.4/2.68 ) = 0.98
That is, ( x - 63.4/2.68 ) = 2.05
--> x = 2.05 * 2.68 + 63.4 = 68.9047                  
b)
P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.674
P( x-u/s.d < x - 63.4/2.68 ) = 0.25
That is, ( x - 63.4/2.68 ) = -0.67
--> x = -0.67 * 2.68 + 63.4 = 61.5937