Question

A menswear manufacturer knows that the height of all men is normal with a mean of 69 inches and a standard deviation of 3 inches.
a) What proportion of all men have a height between 69 and 74 inches?
b) What proportion of all men have a height between 67 and 74 inches?
c) What is the 95th (and 99th) percentile of all men’s heights?

Answer 1

Solution :

(a)

P(69 < x < 74) = P[(69 - 69)/ 3) < (x - ) /  < (74 - 69) / 3) ]

= P(0 < z < 1.67)

= P(z < 1.67) - P(z < 0)

= 0.4525

(b)

P(67 < x < 74) = P[(67 - 69)/ 3) < (x - ) /  < (74 - 69) / 3) ]

= P( -0.67 < z < 1.67)

= P(z < 1.67) - P(z < -0.67)

= 0.7011

(c)

P(Z < 1.645) = 0.95

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 * 3 + 69 = 73.935

95th percentile = 73.935

P(Z < 2.326) = 0.99

z = 2.326

Using z-score formula,

x = z * +

x = 2.326 * 3 + 69 = 75.978

99th percentile = 75.978