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SOLUTIONS TO MIDTERM VERSION 1 1) The random variable X= # Likely Voters for the Democrat (in the sample of size 3) has a binomial distribution with n=3, p=.4. The Democrat is preferred if either 2 or 3 in the sample support the Democrat. So we need 3 3 p(2) p(3) p 2 q p 3 3(.4 2 )(1 .4) .4 3 .352. 2 3 Answer is C. 2) We first need to determine n and p. Since the mean is np = 2 and the variance is npq = 1, we can divide the second equation by the first to conclude that q = .5, and therefore p = .5 and then the first equation implies that n=4. If X is binomial with n=4, p=.5, we need Prob{X ≥ 1} = 1 −Prob{X=0} = 1 .5 4 .9375. Answer is D. 3) If X=Profit, then E[X] = (50000)(.6) + 0(.4) = 30,000. Thus, Var(X)=(50000−30000)2(.6)+(0−30000)2(.4) = 600,000,000 so the standard deviation is the square root of the variance, $24,494.90. Answer is A. 4) They cannot be independent since P(A|B) = P(A B) / P(B) = 0 / P(B) = 0 which is not equal to P(A). Answer is A. 5) Note that σ = 2 and n = 50, so we get x = / n 2 / 50 .2828. Answer is B. 6) Since the variance of the binomial distribution is npq the standard deviation is 5(.4)(.6) 1.095. Answer is D. 7) If X is the normal random variable, then P(X < 0) = P{Z < (0−2)/4} = P{Z < −.5} = .5 − .1915 = .3085. Answer is A. 8) If Xi is the profit for the ith round, we want = Prob (X1 + …+X50 ≥10) = Prob ( X ≥ .2). By the Central Limit Theorem, we can treat X as having a normal distribution with mean x .014 and standard deviation 1.00 = .1414. Thus, we get x n 50 Prob { X ≥ .2}=Prob{Z≥(.2+.014)/.1414}=Prob{Z≥1.51}=.5−.4345=.0655. Answer is D. 9) From Table 5, Prob{Z < −1.5}=.5−.4332 = .0668. Answer is A. 10) If A={Big Change Today} and B={Big Change Tomorrow}, then P(A)=P(B)=.04. From the statement of the problem, we have P(B|A)=.1. Note that A, B are not independent. Using the rule of complements and the general multiplication rule, we conclude that the probability of no big change today and no big change tomorrow is P( A B ) 1 P( A B) 1 [ P( A) P( B) P( A) P( B | A)] 1 [.04 .04 .04(.1)] .924. Answer is D. 11) If X is the average number of fans for the next n=30 days, and Xi is the number of fans for the ith then Xi has mean μ=1.5 and standard deviation σ =1. Thus, X has mean x 1.5 and standard deviation 1 x .1826 . Therefore, by the Central Limit Thorem, n 30 Prob{Win 50 free plays} = Prob{Total number of fans ≥ 40} = Prob{ X ≥1.333} = Prob{Z ≥(1.333−1.5)/.1826} = Prob{Z≥−.91}=.5+.3186=.8186. Now, let Y=Number of Free plays. We have shown that Y will equal 50 with probability .8186, and Y will equal zero with probability 1−.8186 = .1814. Thus, E[Y]=50(.8186)+0 = 40.93. Answer is B. 12) X= #Profitable Years of next 10 has a binomial distribution with n=10, p=.75. We need Prob{X≥9} = 10 10 p(9) p(10) (.759 )(.251 ) (.7510 ) (10)(.759 )(.251 ) .7510 .2440. 9 10 Answer is D. 13) Since the z-score is standard normal, we have Prob(Z>1)=.5−.3413 = .1587. Answer is A. 14) X=Total Number of Heads in 150 tosses is binomial with n=150, p=.5, so the standard deviation of X is npq (150)(.5)(.5) 6.124. Answer is C. 15) The general addition rule is P(AB)=P(A)+P(B)−P(A B). Comparing this with the formula given in the problem, we conclude that P(A B)=P(A)P(B), which can only happen if A and B are independent. Answer is B.