4. Assume the height of women in the U.S. is normally distributed with a mean of

Question

4. Assume the height of women in the U.S. is normally distributed with a mean of 5 feet 4 inches and a standard deviation of 4 inches. (hint: first convert feet to inches) a. How tall would a woman need to be in order to be among the tallest 2.27% of US women? b. If a model needs to be at least 5 feet 8 inches tall, what percentage of females in the U.S. are tall enough to quality as models? c. What percentage of women in the US. are between 5, and 54"'? d. What is the height at 2 standard deviations above the mean? e. What percentage of women in the U.S. are between 5" 4" and 5' 8 5. What is the cumulative percentage at -1 standard deviation? 6. What is the cumulative percentage at 0 standard deviations? 7. What is the cumulative percentage at 2 standard deviations? Normal, Beill-shaped Curve 13.59%| 3413% | 3413%|1359% Percentage of .13% : 2.14% 4% : .13% cases in 8 portions of the curve Standard Deviations -4 -3 -20 -10 440 Cumm 1 0.1% 23% 15.9% 50% 84.1% 97.7% 99.9% 5 10 20 30 40 50 60 70 80 90 959 Normal Curve4 1020304050607080” Equivalents 108

Explanation / Answer

here mean =5*12+4 =64 inch

a) tallest 2.27% will be 2 stdd eviation above mean =64+2*4 =72 inch =6 ft

b) as 5 feet 8 inch is one std deviation abvoe hence (100-84.1) =15.9% are tall enough.

c)

between 5 and 5' 4'' =(50-15.9) =34.1%

d) height =mean +2*std deviation =64+2*4 =72 inch =6 feet

e) % between  5' 4'' and  5' 8'' =84.1-50 =34.1%

5) cumulative % at -1 std deviation =15.9%

6)  cumulative % at 0 std deviation = 50%

7) cumulative % at 2 std deviation =97.7%

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