Question

An article reported that some company was sued in Texas for sex discrimination over a minimum height requirement of 5 ft 6 in. The suit claimed that this restriction eliminated more than 92% of adult females from consideration. Let x represent the height of a randomly selected adult woman. Suppose that x is approximately normally distributed with mean 65 in. (5 ft 5 in.) and standard deviation 2 in.

(a)

Is the claim that 92% of all women are shorter than 5 ft 6 in. correct?

YesNo    

(b)

What proportion of adult women (in percent) would be excluded from employment as a result of the height restriction? (Round your answer to two decimal places.)

%

You may need to use the appropriate table in the appendix or technology to answer this question.

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	65
std deviation   =σ=	5.000

probability =

P(X<66)

=

P(Z<0.2)=

0.5793~ 57.93%

therefore only 57.93 % are shorter than 5 ft 6 in

Is the claim that 92% of all women are shorter than 5 ft 6 in. correct? : NO

b)

t proportion of adult women (in percent) would be excluded from employment as a result of the height restriction =57.93%