document.write( "Question 1088691: A survey found that​ women's heights are normally distributed with mean 63.5 in and standard deviation 2.2 in. A branch of the military requires​ women's heights to be between 58 in and 80 in.
\n" ); document.write( "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?
\n" ); document.write( "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?
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Algebra.Com's Answer #703323 by Boreal(15235)\"\" \"About 
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z value for 58 is (58-63.5/2.2=-5.5/2.2=-2.5
\n" ); document.write( "z value for 80 is (80-63.5)/2.2=16.5/2.2=7.5
\n" ); document.write( "The probability of z being between these is 0.994
\n" ); document.write( "More women will be denied because they are too short. Virtually all will be within the high height limit.
\n" ); document.write( "for the shortest 1% the z value is -2.326
\n" ); document.write( "for the tallest 2%, the z-value is +2.055
\n" ); document.write( "multiply this by the sd to get -5.12, which makes the lower limit 63.5-5.12=58.4 inches lower limit
\n" ); document.write( "and +4.52 above this to get 68.0 inches upper limit (58.4, 68.0)
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