Question

In a survey of women in a certain country​ (ages 20-​29), the mean height was 65.1 inches with a standard deviation of 2.84 inches. Answer the following questions about the specified normal distribution. ​(a) What height represents the 95th ​percentile? ​(b) What height represents the first​ quartile?

Answer 1

Solution:

Given: Women in a certain country​ (ages 20-​29), follow a normal distribution with the mean height was 65.1 inches with a standard deviation of 2.84 inches.

Part a) What height represents the 95^th ​percentile?​

That is find x value such that:

P( X < x ) = 95%

P( X < x ) = 0.95

Thus find z value such that:

P( Z < z ) =0.95

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus z  = 1.645

Now use following formula to find x value:

Thus  the 95^th ​percentile height is 69.77 inches.

Part b) What height represents the first​ quartile?

Area below first quartile is 25%.

That is:

P(X < Q1 ) = 25%

P(X < Q1 ) = 0.25

Thus find z value such that:

P( Z < z ) =0.25

Look in z table for Area = 0.2500 or its closest area and find z value.

Area 0.2514 is closest to 0.2500 and it corresponds to -0.6 and 0.07

Thus z = -0.67

Now use following formula to find Q1 value: