Number 8 only 7. Suppose that a recent poll of America households about pet ownership found that forhouseholds with one pet, 39% owned a dog. Suppose a random sample of 12 households withonly one pet is selected. Use Binomial distribution to find the probability that 3 households willhave a dog? Show work.0.15268. Suppose that a recent poll of American households about pet ownership found that forhouseholds with one pet, 39% owned a dog. Suppose a random sample of 120 households withonly one pet is selected.(a) Find the usual values interval for households with a pet dog.(41.5, 52.4)(b) Determine whether it is unusual to find 40 household with a pet dog in a random,independent sample of 120 households.9. Two balls are drawn successively without replacement from a bucket containing four red andthree blue balls. Assume that all balls are equally likely to be chosen.a) Find the probability that both balls are red.b) Find the probability that at least one ball is blue.c) Find the probability that first ball is blue and the second one is red.DesktopDELLE5F6F7F8F9F10F11F12PrtScr近

Given information: P(Owned dog) = 0.39 n = 120a) Use the excel formula, "=NORM.S.INV(0.975)" for determining the critical z-score. z0.95 = 1.96 C.I=p^±zα/2×p^(1-p^)n=0.39±1.96×0.39(1-0.39)120=0.39±1.96×0.044525=0.39±0.0873C.I=(0.3027, 0.4773) Usual values interval for households with a pet dog=1200.3027,0.4773=120·0.3027,120·0.4773=36.324,57.276≈(36, 57)b) 95% confidence interval for households with a pet dog is C.I = (36,57). Since the 95% confidence interval contains 40 households with a pet dog, hence it is not unusual to find 40 household with a pet dog in a random independent sample of 120 households.