Question

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.

LOADING...

Click to view page 2 of the table.

LOADING...

(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.)

Answer #1

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

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