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The U.S. Army requires women\'s heights to between 58 in. and 75 in. Women\'s he

ID: 3233292 • Letter: T

Question

The U.S. Army requires women's heights to between 58 in. and 75 in. Women's heights (in USA) are normally distributed with a mean of 63.8 in. and a standard deviation of 2.6 in. Step 1: Determine the 2 scores for each of the Army women's height requirements (58" & 75"). Use Excel, fx, Statistical, and STANDARDIZE option to determine the z score. Z (58")= Z(75") = Step 2: Determine the probability of finding each women's height using Excel, fx, Statistical, NORM DIST.(x = 58, x = 63.8, s = 2.6, cumulative = 1) P(58") = P(75") =. Step 3: Determine the percentage of women meeting the height requirements of the U.S. Army P(75") -P(58") =. Step 4: Are many women being denied the opportunity to join the Army because they are too short or tall? Yes or No? Part B: The Boeing 757 doorway is 72 inches. Men's heights are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. Women's heights are normally distributed with a mean of 63.8 in. and a standard deviation of 2.6 in. Step 1: Determine the z scores for men and women at a height of 72". Use Excel, fx, Statistical, and STANDARDIZE option to determine the z-score. Z (men) = Z(women) = Step 2: What percentage of men women can fit through the doorway without bending? Use Excel, fx, Statistical, NORMDIST (cumulative = 1) to determine the probability. P(men) = (women) = Step 3: Determine the percentage of men and women too tall to fit through a standard doorway. % men too tall % women too tall Step 4: Are there many men or women too tall to fit through a typical aircraft doorway? Yes or No?

Explanation / Answer

z=(X-mean)/sd=(X-63.8)/2.6

step1. for X=58, z=(58-63.8)/2.6=-2.23

for X=75, z=(75-63.8)/2.6=4.31

Step2.P(58)=0.0128, P(75)=0.9999

step3.P(75)-P(58)=0.9999-0.128=0.872

step.No. , as 87.2% were allowed only 12.8% are denied

Part(B)

step1, Z(men)=(72-69)/2.9=1.0344

Z(women)=(72-63.8)/2.6=3.1538

step2:P(men)=0.8495=84.95%, P(women)-0.9992=99.92%

step3:%of men=100-84.95=15.05

%women=100-99.92=0.08%

step4. yes,

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