Question

451 In the United States, the mean and standard deviation of adult women's heights are 65 inches (5 feet 5 inches) and 3.5 inches, respectively. Suppose the American adult women's heights have a normal distribution. a. If a woman is selected at random in the United States, find the probability that she is taller than 5 feet 8 inches. b. Find the 72nd percentile of the distribution of heights of American women. c. If 100 women are selected at random in the United States, find an approximate probability that exactly 20 of them are taller than 5 feet 8 inches.

Answer 1

Solution 8 base of from given datas Mean (m): 65 standard deviation (6): 86 las find the probability Herer find the probability that she is taller than 5 feel. 8 Inches? - 5 feel. 8 Suches One feet = 12 inches son 5X1200 60+8= 68 Inches son P(x > 08 -2 P(x368) - PC & League, *745) =P(Z > 98765 =P(7225) = P( Zz 0.8571) :1-P (740.8570) The = 1-0.8043 TP(5>68) = 0.1957 . value from normal distribution table ... So, the required probability is 0.1957

(b) find the 72nd percentile of the distribution of heights of American women P): 727 Zoritical value = 7072: 0.5828 Plaç x): 0.72 Plz Plze ) = 0.5828 ) 30-5828 X-65 -0.5828 35 À -65 = 2.0398 67.0398 ~67.04 = 2.0398+65 a = 67.04 ) I C) fonom partias the probability p=0.1957 wing the women taller than sheet8 Jucher? Criven n = {CÓ X: 20 / P: Q• Sở 1 Ptv - V1.0.1957 90.8043 The binomial distribution formula is given by PIX=x) = (culpju (9,467 P(x=20) = (10.20) (0-1957.90 (0-8043) 700-20 = {(53598 xv°°) (6.789.2 X10-15) (0.0128)] Pléxactly 20) = 0.0987