Question

In a survey of women in a certain country​ (ages 20minus−​29), the mean height was 65.6...

In a survey of women in a certain country​ (ages

20minus−​29),

the mean height was

65.6

inches with a standard deviation of

2.81

inches. Answer the following questions about the specified normal distribution.​(a) What height represents the

95th

​percentile?

​(b) What height represents the first​ quartile?

Click to view page 1 of the table.

Click to view page 2 of the table.

​(a) The height that represents the

95th

percentile is

nothing

inches.

​(Round to two decimal places as​ needed.)

​(b) The height that represents the first quartile is

nothing

inches.

​(Round to two decimal places as​ needed.)

Solution :

mean = = 65.6

standard deviation = = 2.81

Using standard normal table,

(a)

P(Z < z) = 95%

P(Z < 1.645) = 0.95

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 * 2.81 + 65.6 = 70.22

95th percentile = 70.22

(b)

P(Z < z) = 25%

P(Z < -0.67) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 2.81 + 65.6 = 63.72

First quartile = 63.72