Question

The heights of all adult American women are normally distributed with a mean of 63.6 inches and a standard deviation of 5 inches. Give the standard (z) score and approximate percentile (from the tables) for women with each of the following heights:

1. 63 inches

2. 60 inches

3. 62.5 inches

Answer 1

Solution :

Given that ,

mean = = 63.6

standard deviation = = 5

a) x = 63

Using z-score formula,

z = x - /   

z = 63 - 63.6 / 5

standard score = z = -0.12

= P(z < -0.12)

Using z table,

= 0.4522

The percentage is = 45.22%

percentile is = 45 th

b) x = 60

Using z-score formula,

z = x - /   

z = 60 - 63.6 / 5

standard score = z = -0.72

= P(z < -0.72)

Using z table,

= 0.2358

The percentage is = 23.58%

percentile is = 24 th

c) x = 62.5

Using z-score formula,

z = x - /   

z = 62.5 - 63.6 / 5

standard score = z = -0.22

= P(z < -0.22)

Using z table,

= 0.4129

The percentage is = 41.29%

percentile is = 41 th