Question

A survey found that​ women's heights are normally distributed with mean 63.9in. and standard deviation 3.1 in. The survey also found that​ men's heights are normally distributed with mean 68.7 in. and standard deviation 3.7 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63in. Complete parts​ (a) and​ (b) below.

The percentage of men who meet the height requirement ?

If the heigh 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

Given:

For women :

Mean = = 63.9

Standard deviation = = 3.1

Let X be the women's heights are normally distributed.

For men :

Mean = = 68.7

Standard deviation = = 3.7

Let X be the men's heights are normally distributed

Ansa:

The percentage of men who meet the height requirement ?

ie P(57<x<63)

For men :

Mean = = 68.7

Standard deviation = = 3.7

P(57<x<63)=P((57-68.7)/3.7<(x-)/<(63-68.7)*/3.7)

=P(-3.162<z<-1.5405)

=P(z<-1.5406)-P(z<-3.162)

=0.06171-0.0008

=0.0609

# 6.09% of men who meet the height requirement

Ansb:

P(z<z)0.05

P(z<-1.645)=0.05

z=-1.645

from z-score formula.

x1=+z*

=68.7+(-1.645*3.7)

x1=62.61

P(z>0)=0.5

ie z=0

from z-score formula.

x2=+z*

=68.7+(0*3.7)

=68.7

x2=68.7