A survey found that womens heights are normally distributed

A survey found that women\'s heights are normally distributed with mean 63.4 in. and standard deviation 2.3 in. The survey also found that men\'s heights are normally distributed with a mean 68.7 in. 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 3 in. Find the percentage of women meeting the height requirement. (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 the are the new height requirements? at least ___ at most ____

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 =    75      
u = mean =    63.4      
          
s = standard deviation =    2.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.782608696      
z2 = upper z score = (x2 - u) / s =    5.043478261      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.00269619      
P(z < z2) =    0.999999771      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.997303581 or 99.73% [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 =    75      
u = mean =    68.7      
          
s = standard deviation =    2.9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -4.034482759      
z2 = upper z score = (x2 - u) / s =    2.172413793      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    2.73614E-05      
P(z < z2) =    0.985087767      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.985060406 or 98.51% [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 = 1 - 0.05 =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    68.7      
z = the critical z score =    1.644853627      
s = standard deviation =    2.9      
          
Then          
          
x = upper critical value =    73.47007552   [AT MOST]

  
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For the 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 =    63.4      
z = the critical z score =    -1.644853627      
s = standard deviation =    2.3      
          
Then          
          
x = critical value =    59.61683666   [AT LEAST]

Thus, it is AT LEAST 59.61683666 AND AT MOST 73.47007552. [ANSWER]