A survey found that womens heights are normally distributed

A survey found that women\'s heights are normally distributed with mean 62.8 in. and standard deviation 2.6in. The survey also found that men\'s heights are normally distributed with a mean 68.7 in. and standard deviation 2.9.

a. . Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 99in. and a maximum of 6 ft 33in. Find the percentage of women meeting the height requirement.

The percentage of women who meet the height requirement is _________

b. Find the percentage of men meeting the height requirement

The percentage of men who meet the height requirement is ____________ (Round to two decimal places as needed.)

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

The new height requirements are at ________ in. and at most ________ in.(Round to one decimal place as needed.)

Solution

Normal Distribution
Mean ( u ) =62.8
Standard Deviation ( sd )=2.6
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 59.88) = (59.88-62.8)/2.6
= -2.92/2.6 = -1.1231
= P ( Z <-1.1231) From Standard Normal Table
= 0.1307
P(X < 75.96) = (75.96-62.8)/2.6
= 13.16/2.6 = 5.0615
= P ( Z <5.0615) From Standard Normal Table
= 1
P(59.88 < X < 75.96) = 1-0.1307 = 0.8693                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 59.88) = (59.88-68.7)/2.9
= -8.82/2.9 = -3.0414
= P ( Z <-3.0414) From Standard Normal Table
= 0.00118
P(X < 75.96) = (75.96-68.7)/2.9
= 7.26/2.9 = 2.5034
= P ( Z <2.5034) From Standard Normal Table
= 0.99385
P(59.88 < X < 75.96) = 0.99385-0.00118 = 0.9927                  
c)
Tallest 5% of men
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.64
P( x-u/ (s.d) > x - 68.7/2.9) = 0.05
That is, ( x - 68.7/2.9) = 1.64
--> x = 1.64 * 2.9+68.7 = 73.4705                  

Smallest 5% of women
P ( Z < x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.645
P( x-u/s.d < x - 62.8/2.6 ) = 0.05
That is, ( x - 62.8/2.6 ) = -1.64
--> x = -1.64 * 2.6 + 62.8 = 58.523                  

The new height requirements are at 58.5 in. and at most 73.5 in.