A survey found that womens heights are normally distributed

A survey found that women\'s heights are normally distributed with mean

62.262.2

in. and standard deviation

2.42.4

in. The survey also found that men\'s heights are normally distributed with a mean

67.367.3

in. and standard deviation

2.92.9.

Complete parts a through c below.

a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft

88

in. and a maximum of 6 ft

33

in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is what percent?

B. Find the percentage of men meeting the height requirement.

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 least
? in. and at most
? in.

Solution

A)

Z*_Upper = (75 - 62.26)/2.42 = 5.26
Z*_Lower = (56 - 62.26)/2.42 = -2.59


The requirement is to get p(-2.59 < Z < 5.26) = p(Z<5.26) - p(Z<-2.59)

= 0.9952 ~ 99.52 % Answer

B)

Z*_Upper = (75 - 67.37)/2.92 = 2.61
Z*_Lower = (56 - 67.37)/2.92 = -3.89


The requirement is to get p(-3.89 < Z < 2.61) = p(Z<2.61) - p(Z<-3.89)

= 0.9954 ~ 99.54 % Answer

C)

Now we know the area 5% or 0.05 that we find the z value so we can manipulate our formula

z = (x – µ) / and solve for x to get x = µ + z

At 5% or 0.05 look at the four digit values (as this is the area) and I find the z value of -1.64. So I have the lower height as

x = 62.26 + (-1.64)(2.42) = 58.29 Answer

Upper height = 67.37 + (-1.64)(2.92) = 62.58 Answer