Mean and standard deviation

Question

a survey found that women’s heights are normally distributed with mean 62.7 in. and standard deviation 2.3in. The survey also found that men’s heights are normally distributed with a mean 67.5 in. And standard deviation 2.8

Most of live characters at an amusement park have height requirements with a minimum of 4ft 9 in. And a maximum of 6ft 3 in. Find the percentage of women meeting the height requirement. 

The percentage of women who meet the height requirements is ____% 

Answer 1

Convert height requirements to inches

 

4'9" = 4.75 feet

 

4.75 feet x 12 = 57 inches

 

6'3" = 6.25 feet

 

6.25 feet = 75 inches

 

Now use these measurements to find Z values

 

(57 - 62.7)/2.3 = -2.48

 

(62.7 - 75)/2.3 = 5.35

 

Z value for -2.48 = .0066

 

Z value for 5.35 > .9999

 

P(-2.48 < x < 5.35) = (1 - .0066)

 

Percentage for the women = .9934 = 99.34%

 

(57 - 67.5)/2.8 = -3.75

 

(75 - 67.5/2.8 = 2.68

 

Z value for -3.75 = approx .9999

 

Z value for 2.68 = .9962

 

P(-3.75 < x < 2.67) = (.9962 - .0001)

 

Percentage for the men = .9961 = 99.61%