A survey found that women\'s heights are normally distributed with mean 63.4 in

Question

A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.5 in. A branch of the military requires women's heights to the between 50 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being derived the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible expect the shortest 1% and the talest 2%, what are the new heigh requirements? a. The percentage of women who meet the height requirements is % (Found to two decimal places ans needed.)

Explanation / Answer

A)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    58      
x2 = upper bound =    80      
u = mean =    63.4      
          
s = standard deviation =    2.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.16      
z2 = upper z score = (x2 - u) / s =    6.64      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.0154      
P(z < z2) =    1      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.9846      

Thus, those outside this interval is the complement =    0.0154 = 1.54% [ANSWER]

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B)

FOR SHORTEST 1%:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.01      
          
Then, using table or technology,          
          
z =    -2.33      
          
As x = u + z * s,          
          
where          
          
u = mean =    63.4      
z = the critical z score =    -2.33      
s = standard deviation =    2.5      
          
Then          
          
x = critical value =    57.575   [ANSWER, SHORTEST 1%]

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FOR TALLEST 2%:

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1-0.02 =   0.98      
          
Then, using table or technology,          
          
z =    2.05      
          
As x = u + z * s,          
          
where          
          
u = mean =    63.4      
z = the critical z score =    2.05      
s = standard deviation =    2.5      
          
Then          
          
x = critical value =    68.525   [ANSWER, TALLEST 2%]  
  
      

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