Question

Homework: Homework: Chapter5 (Due=Mar01) Score: 0 of 1 pt < 31 of 64 (54 complete) HW Score: 71,09%, 45.5 of 64 pts 5.3.31 Question Help In a survey of women in a certain country (ages 20-29). the mean height was 68.3 inches with a standard deviation of 2.77 inches. Answer the following questions about the specified normal distribution (a) What height represents the 5th 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 5th percentile is (Round to two decimal places as needed.) inches.

Answer 1

Solution:-

Given that,

mean = $\mu$ = 66.3

standard deviation = $\sigma$ = 2.77

Using standard normal table,

a ) P(Z < z) = 95%

= P(Z < z) = 0.95  

= P(Z < 1.645 ) = 0.95

z = 1.64

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.64 * 2.77 + 66.3

x = 70.8428

95^th Percentile = 70.84

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.67

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.67 * 2.77 + 66.3

x = 64.4441

25^th Percentile = 64.44