Question

in a survey of men in a certain country ages (20-29) the mean height was 64.6 inches with a standard deviation of 2.8 inches.

1. the height that represents the 99th percentile is ?? inches (round to two decimals places as needed)

2. the height that represents the first quartile is ?? inches (round to two decimal places as needed)

Answer 1

Given that,

mean = = 64.6

standard deviation = = 2.8

Using standard normal table,

P(Z < z) =99 %

= P(Z < z) = 0.99  

= P(Z < 2.33) = 0.99

z =2.33 Using standard normal z table,

Using z-score formula  

x= z * +

x= 2.33 *2.8+64.6

x= 71.124

x=71.12

(B)

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.67 ) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 2.8+ 64.6

x = 62.724

First quartile =Q1 = 62.72