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A survey found that womens heights are normally distributed with a mean 63.6 in and a standard deviation 2.5 in. A branch of military requries womens heights to be between 58 in and 80 in a) find the percentage of women meeting the height requirement. are too many women being denied the opportunity to join this branch of the millitary because they are too short or too tall b) if the branch of millitary changes the height requirements so that all women are eligable except the shortest 1% and the tallest 2% what are the new height requirements

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Answer:

Step-by-step explanation:

We shall calculate the standard variate factor (Z factor) for each of the two given values

[tex]Z=\frac{X-\bar{X}}{\sigma }[/tex] Thus

[tex]Z_{1}=\frac{X-\bar{X}}{\sigma }\\\\Z_{1}=\frac{58-63.6}{2.5}\\\\Z_{1}=-2.24[/tex]

Similarly

[tex]Z_{2}=\frac{80-63.6}{2.5}\\\\Z_{2}=6.56[/tex]

Using the standard tables we get the  percentage of area between [tex]Z_{1},Z_{2}[/tex] is [tex]98.75%[/tex]

thus 98.75% women are getting the job

b)

No too many women are not denied the opportunity of the job As 98.75% women get the job.

c)

Using the new requirements the area corresponding to the 1% shortest is

The height requirements are height should be greater than 57.78 inches and it should not be more than 68.753 inches corresponding to the tallest 2%. These values are arrived at using the standard normal distribution table values.