UtatistiCS FIial Examination Time Remalining: 01:45.25 Subm This Question: 1 pt
ID: 3266203 • Letter: U
Question
UtatistiCS FIial Examination Time Remalining: 01:45.25 Subm This Question: 1 pt 40f 25(1 complete) This Test: 25 pts po A survey found that women's a. Find the percentage of women meeting the height requirement. Are many women being b. If this branch of the military changes the height requirements so that all women are heights are normally distributed with mean 63.9 in and standard deviation 2.3 in. A branch of the miltary requires women's heights to be between 58 in and 80 in. denied the opportunity to join this branch ofthe military because they are too short too tar? eligie except the shortest 1% and the talest 2%, what are the new height requrements? Click to view page 1 of the table, Click to view page 2 of the table a, The percentage of women who meet the height requirement is%. (Round to two decimal places as needed.) Are many women being denied the opportunity to join this branch of the military because they are too short or too tail? A OB, ° C. 0 D. , because the percentage of women who meet the height requirement is fairly small , because only a small percentage of women are not alowed tojoin this branch of the military because of their height Yes, because the percentage of women who meet the height requirement is fairly large Yes, because a large percentage of women are not allowed to join this branch of the military because of their height b. For the new height requirements, this branch of the military requires women's heights to be at leastin and at mostin (Round to one decimal place as needed.) Click to select your answerfs). 8 9
Explanation / Answer
Mean is 63.9 and standard deviation is 2.3
Upper value is 80, thus z value for this is (80-63.9)/2.3 =7, Thus the area below this is 1
Lower value is 58, thus z is (58-63.9)/2.3 =-2.57, thus from normal table we get p=0.005
Thus area between these values is 1-0.005=0.995
Answer B is correct
Shortest 1% means z value of -2.33 (from table), thus lower limit is 63.9-2.33*2.3=58.54
Tallest 2% means z value for 98% area to left which is 2.06 , thus upper limit is 63.9-2.06*2.3= 59.16
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