A survey found that womens heights are normally distributed

A survey found that women\'s heights are normally distributed with mean 62.4in. and standard deviation 2.9 in. The survey also found that men\'s heights are normally distributed with a mean 68.5in. and standard deviation 2.9.

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 9 in. and a maximum of 6 ft 4 in. Find the percentage of women meeting the height requirement.

The percentage of women who meet the height requirement is _________

(Round to two decimal places as needed.)

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 least _________ in. and at most _________in.

(Round to one decimal place as needed.)

Solution

A)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    57      
x2 = upper bound =    76      
u = mean =    62.4      
          
s = standard deviation =    2.9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.862068966      
z2 = upper z score = (x2 - u) / s =    4.689655172      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.031296685      
P(z < z2) =    0.999998632      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.968701947 OR 96.87% [answer]

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B)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    57      
x2 = upper bound =    76      
u = mean =    68.5      
          
s = standard deviation =    2.9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3.965517241      
z2 = upper z score = (x2 - u) / s =    2.586206897      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    3.66185E-05      
P(z < z2) =    0.995148067      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.995111448 or 99.51% [ANSWER]
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c)

For tallest 5% of men:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    68.5      
z = the critical z score =    1.644853627      
s = standard deviation =    2.9      
          
Then          
          
x = critical value =    73.27007552   [ANSWER]

*****

For shortest 5% of women:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.05      
          
Then, using table or technology,          
          
z =    -1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    62.4      
z = the critical z score =    -1.644853627      
s = standard deviation =    2.9      
          
Then          
          
x = critical value =    57.62992448   [ANSWER]

Thus, the new height requirements are at least 57.6 in and at most 73.3 in. [ANSWER]