Question

6. In a survey of men in the U.S. (aged 20 to 29), the mean height is 68.7 inches with a standard deviation of 3.1 inches. Assume this height data is normally distributed.

a. What percentage of these men are taller than 72 inches?

b. Find the 45th percentile. Interpret this value in a sentence.

c. How tall are the middle 95% of men?

Answer 1

Solution :

Given that,

mean = = 68.7

standard deviation = = 3.1

a ) P (x >72 )

= 1 - P (x < 72 )

= 1 - P ( x -  / ) < ( 72 - 68.7 / 3.1)

= 1 - P ( z < 3.3 / 3.1 )

= 1 - P ( z < 1.06 )

Using z table

= 1 - 0.8554

= 0.1446

Probability = 0.1446

P(Z < z) = 45%

P(Z < z) = 0.45

P(Z < - 0.61) = 0.45

z = - 0.13

Using z-score formula,

x = z * +

x = - 0.13 * 3.1 + 68.7

x = 68.30

c ) P(-z < Z < z) = 95%
P(Z < z) - P(Z < z) = 0.95
2P(Z < z) - 1 = 0.95
2P(Z < z ) = 1 + 0.95
2P(Z < z) = 1.95
P(Z < z) = 1.95 / 2
P(Z < z) = 0.975
z = -1.96 and z = 1.96

Using z-score formula,

x = z * +

x = - 1.96 * 3.1 + 68.7

x = 62.62

Using z-score formula,

x = z * +

x = 1.96 * 3.1 + 68.7

x = 74.78