SOLUTION: Hello can you please assist me in answering this question.
Men's heights are normally distributed with mean 59.0 and standard eviation 2.8in.
Question: Marine Corps heigt req
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Question 564035: Hello can you please assist me in answering this question.
Men's heights are normally distributed with mean 59.0 and standard eviation 2.8in.
Question: Marine Corps heigt requirement for men. The US Marine Corps requires that men have heights between 64 in. and 80 in.
(a) Find the percentage of men who meet the height requirements. Are many men denied the opportunity to become a Marine because they do not satisfy the height requirement.
(b) If the height requirements are changed so that all men are eligible except the shortest 3% and the tallest 4%, what are the new height requirements?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Men's heights are normally distributed with
mean 59.0 and standard deviation 2.8in.
---------------------------------------------
Question: Marine Corps height requirement for men. The US Marine Corps requires that men have heights between 64 in. and 80 in.
---------------------------------------------
(a) Find the percentage of men who meet the height requirements.
z(64) = (64-59)/2.8 = 1.7857
z(80) = (80-59)/2.8 = 7.5
P(64 <= x <= 80) = P(1.78457<= z <=7.5) = 0.0371 = 3.71%
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Are many men denied the opportunity to become a Marine because they do not satisfy the height requirement.
---
Since less than 4% meet the height requirements many men must be denied
entrance.
================
(b) If the height requirements are changed so that all men are eligible except the shortest 3% and the tallest 4%, what are the new height requirements?
Find the z-value with a 3% left tail: invNorm(0.03) = -1.88
Newer lower requirement would be -1.88*2.8 + 59 = 53.73 inches
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Find the z-value with a left tail of 96%: invNorm(0.96) = 1.75
Newer upper requirement would be 1.75*2.8 + 59 = 63.90 inches
------------------------
Cheers,
Stan H.
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