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A survey found that women's heights are normally distributed with mean 63.5 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? 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?
A survey found that women's heights are normally distributed with mean 63.5 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? 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?
The survey follows ofwomen's heightanormal distribution.Theheightof 98.51% ofwomenthat meet theheight requirementare between58 inches and 80 inches.The newheight requirementswould be 57.7 to 68.6 inchesThe given parameters are:--- mean--- standard deviation(a) Percentage of women between 58 and 80 inchesThis means that: x = 58 and x = 80When x = 58, thez-scoreis:This givesWhen x = 80, thez-scoreis:So, thepercentageofwomenis:Substitute known valuesUsing thep-value table, we have:Express as percentageApproximateThis means that:Theheightof 98.51% ofwomenthat meet theheight requirementare between58 inches and 80 inches.So,many women(outside this range) would bedeniedtheopportunity, because they are eithertoo shortortoo tall.(b) Change of requirementShortest = 1%Tallest = 2%If thetallestis 2%, then theupper endof theshortest rangeis 98% (i.e. 100% - 2%).So, we have:Shortest = 1% to 98%This means that:Thep valuesare: 1% to 98%Using thez-score tableWhen p = 1%, z = -2.32635When p = 98%, z = 2.05375Next, we calculate the x values fromSubstituteMultiply through by 2.5Make x the subjectApproximateSimilarly, substituteinMultiply through by 2.5Make x the subjectApproximateHence, the newheight requirementswould be 57.7 to 68.6 inchesRead more aboutprobabilities of normal distributionsat:brainly.com/question/6476990...