Question

A survey found that​ women's heights are normally distributed with mean 63.4 in. and standard deviation 2.5 in. The survey also found that​ men's heights are normally distributed with mean 68.6 in. and standard deviation 3.3 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 56 in. and a maximum of 64 in. Complete parts​ (a) and​ (b) below.

a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement​ park?

b. If the height requirements are changed to exclude only the tallest​ 50% of men and the shortest​ 5% of​ men, what are the new height​ requirements?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	68.6
std deviation   =σ=	3.3000

  percentage of men meeting the height requirement.:

probability =

P(56<X<64)

=

P(-3.82<Z<-1.39)=

0.0823-0.0001=

0.0822~ 8.22%

(try 8.16 % if above comes wrong due to rounding error)

b)

5th pecentile ; critical z value = -1.645

therefore corresponding value=mean+z*std deviation=

63.17

for 50th percentile is at mean ; therefore corresponding value =68.6

hence new height​ requirements are from 63.2 to 68.6 inch