The average height of a woman aged 20Solution The average he

The average height of a woman aged 20

Solution

The average height of a woman aged 20 - 74 years is 64 inches in 2002, with an increase of approximately one inch from 1960. Suppose the height of a woman is normally distributed with a standard deviation of 2 inches. A) what is the probability that a randomly selected woman in this population is between 58 and 70 inches? Z = 58-64/2 = -3 P(Z < -3) = .0013 Z = 70-64/2 = 3 P(Z < 3) = .9987 P(58 < X < 70) = P(-3 < Z < 3) = P(Z < 3) – P(Z < -3) = .9987-.0013 = .99742 Answer: .9974 B) what are the quartiles of this distribution? Q1: Z score corresponding to 25th percentile = -.6745 -.6745*2+64= 62.65 Q1 = 62.65 Q2 or median: Z score corresponding to 50th percentile = 0 0*2+64 = 64 Q2 = 64 Q3: Z score corresponding to 75th percentile = .6745 .6745*2+64 = 65.35 C) determine the height that is summetric about the mean that includes 90% of this population. This questions sounds like you wanted a z score. If you want the Z score, it is 1.64485 and -1.64485. If you want the heights, this would be 3.2897 inches above and 3.2897 inches below, or 60.71 and 67.29 D) what is the probability that 5 women selected at random from this population all exceed 68 inches? Probability of exceeding 68 inches: Z score = 68-64/2 = 2. P(Z >2) =1 – P(Z <2) = 1 - .9772 = .0228 Probability 5 random women exceeding 68 inches = .0228^5 = 0.00000000609