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 be between 58 in and 80 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 be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? The percentage of women who meet the height requirement is %? (Round to two decimals places as needed).
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

PsyDAG · Answer

a. Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores. Multiply by 100 to get %.

b. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.01 and .02) to get the Z scores. Insert in equation above.

Anonymous · Answer

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bob · Answer

98.75 was the answer my hw accepted