Question

The heights of adult men in America are normally distributed, with a mean of 69.5 inches and a standard deviation of 2.68 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.4 inches and a standard deviation of 2.53 inches.

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?


b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.



c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?



d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

Answer 1

Answer:

Given,

Here let us assume that X be the height of adult men in america

X N(1 , 1)

mean = 1 = 69.5

Standard deviation = 1 = 2.68

let us assume that Y be the height of adult women in america

Y N(2 , 2)

mean = 2 = 64.4

Standard deviation = 2 = 2.53

a)

To determine the z score

Here Xbar = 6feet 3 inches

= (6*12) + 3

= 72 + 3

Xbar = 75

z = (Xbar - 1) / 1

substitute values

z = (75 - 69.5)/2.68

= 5.5/2.68

z = 2.05

b)

To give the percentage of men are SHORTER than 6 feet 3 inches

P(x < 75) = P(z < 2.05)

= 0.9798178 [since from z table]

= 97.98% or 98%

so 97.98% or 98% of men are shorter than 6 feet 3 inches

c)

To give z score

Ybar = 5 feet 11 inches

= (5*12) + 11

= 60 + 11

Ybar = 71

z = (Ybar - 2)/2

= (71 - 64.4)/2.53

= 6.6/2.53

= 2.6086957

z = 2.61

d)

To give percentage of women are TALLER than 5 feet 11 inches

P(Y > 71) = P(z > 2.61)

= 0.0045271 [since from z table]

= 0.0045

So 0.4% of women are TALLER than 5 feet 11 inches