Question

6. Heights in inches for American males aged 20 and over are approximately normally distributed (symmetric) with the mean height 69.3 inches and std deviation 2.99 inches.

a.What percentage of American males in the above age group who are 6 feet or taller?

b.Find the 99th percentile of the American males in the above age group and interpret it.

c. Suppose a random sample of 100 American males aged 20 or more is taken, what is the probability that the sample mean height exceeds 72 inches?

d. Find the 95th percentile for the mean height value for this sample of 100 American males aged 20 or more

Answer 1

Solution :

Given that ,

mean = = 69.3

standard deviation = = 2.99

a) 6 feet = 72 inches.

P(x 72 ) = 1 - P(x   72)

= 1 - P[(x - ) / (72 - 69.3) /2.99 ]

= 1 -  P(z 0.90)  

= 1 - 0.8159 = 0.1841

Probability = 0.1841

b) 99 th percentile is

P(Z < ) = 0.99

z =2.326

Using z-score formula,

x = z * +

x = 2.326 * 2.99 +69.3

x = 76.25

The 99th percentile of the American males in the above age group IS 76.25 .

C)

n = 100

=   = 69.3

= / n = 2.99/ 100 = 0.299

P( > 72 ) = 1 - P( < 72)

= 1 - P[( - ) / < (72-69.3) /0.299 ]

= 1 - P(z < 9.03)

= 1- 1 = 0

Probability = 0.0000

D) 95 th percentile

P(Z < z ) = 0 95

z = 1.645

= z * +   = 1.645*0.299 + 69.3

= 69.79