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 :

Given that ,

mean = = 69

standard deviation = = 3

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

= P( 0< Z < 1.67)

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

Using z table

= 0.9525-0.5

proportion = 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 )

Using z table

= 0.9525-0.2514

proportion = 0.7011

c.

P(Z < z) = 95%

= P(Z < z) = 0.95

z = 1.65 Using standard normal z table,

Using z-score formula

x= z * +

x= 1.65*3+69

x= 73.95

= P(Z < z) = 0.99

z = 2.33 Using standard normal z table,

Using z-score formula

x= z * +

x= 2.33*3+69

x= 75.99

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