Question

According to a survey of 25,000 households. 49% of the people watching a special event on TV enoyed the commercials more than the event tsell. Complete parts a through e below based on a random sample of nine people who watched the special event a. What is the probability that exactly three people enjoyed the commercials more than the event? The probability is (Round to four decimal places as needed) b. What is the probability that less than four people enjoyed the commercials more than the event? The probability is (Round to four decimal places as needed) c. What is the probability that exactly five or six people enjoyed the commercials more than the event? The probability is (Round to four decimal places as needed.) d. What are the mean and standard deviation for this distribution? The meanis (Type an integer or a decimal) The standard deviation is Round to four decimal places as needed) e Construct a histogram for this distribution Choose the correct grach below Click to select your answers)

Answer 1

Given that n=9, P=0.49 using binomial distribution P(x) = C P q n-* (a) P(x= 3) = a. (0.49)}(1-0.49) 9-3 = 0.17389509573 [P(x=3): 0.1739 (6) P(x + 4) = P(x=0) + ((x = 1) t .. & P(x=3) - % (0.4939 (-0.49)9°+ --- --- + ,00.49)*(1-0.499-3 = 0.27398201381 P(x<4)= 0.2740 (c) P(x=500 x=6) = P(x=5) + P(x=6) = 9, (0.49 (1-0.49) + ac (0.49) ° (1-0.49 70 = 0.24078623146 + 0.15422908945 SP (x = 500 X = 6) = 0.3950 (d) Mean = nxp = 9x0.49 = (4.41 S.D = Inpg = Saxo.49×(1-0.49) 11.4997 1 (e) - Answers: optron (A)