Math 425 (Fall 08) Solutions Midterm 2 November 6, 2008


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1 Math 425 (Fall 8) Solutions Midterm 2 November 6, 28 (5 pts) Compute E[X] and Var[X] for i) X a random variable that takes the values, 2, 3 with probabilities.2,.5,.3; ii) X a random variable with the density function given by { x f(x) = 2 if x [, 2] else iii) X a random variable with the cumulative distribution function given by if x < F (x) = ln x if x [, e] if x > e Answer: i) E[X] = = 2. For the variance: ii) E[X 2 ] = = 4.9 Var[X] = E[X 2 ] E[X] 2 = 4.9 (2.) 2 =.49 E[X] = E[X 2 ] = 2 2 x f(x) dx = 2 2 x 2 f(x) dx = 2 2 x 2 dx = x3 6 Var[X] = E[X 2 ] E[X] 2 = = 2 9 iii) You have to compute the density function first x 3 dx = x4 8 2 = = 2 for x [, e]. Thus f(x) = (F (x)) = x Similarly, E[X] = e x f(x) dx = e Var[X] = 2(e ) dx = e
2 2 ( pts) A man claims to have extrasensory perception (ESP). As a test, a fair coin is flipped times, and the man is asked to predict the outcome in advance. He gets out of correct. What is the probability that he would have done at least this well if he had no ESP? ( ) Answer: Binomial distribution B, 2. The requested probability is: P (B(,.5) ) = 2 (( ) ( )) =
3 3 (5 pts) a) Each of the members of a seven judge panel will independently make a correct decision with probability.. If the panel s decision is made by majority rule, what is the probability that the panel makes the correct decision? Answer: Binomial distribution. P (B(,.) 4) = ( ) (.) 4 (.3) ( ) (.) b) On average, 5.2 hurricanes hit a certain region in a year. What is the probability that there will be 3 or fewer hurricanes hitting this year? Answer: Poisson distribution with λ = 5.2. ( P (P oisson(5.2) 3) = e (5.2)2 2! ) + (5.2)3 3! c) A casino patron will continue to make $5 bets on red in roulette until she has won 4 of these bets. What is the probability that she places a total of 9 bets. Answer: Let X be the number of bets until you get the 4th win. The distribution of X is called the negative binomial distribution (with parameters p = 2 and r = 4). We want to compute: ( ) ( ) 9 4 ( P (X = 9) = ( ) = 4 2 2)
4 4 ( pts) What is the probability that out of poker games, you get at least once a straight flush? Note: Straight flush means 5 consecutive cards of the same suit (e.g. 6,,8,9, of diamond). You get 5 cards once and you are not allowed to exchange any. Answer: Binomial distribution/approximation of the binomial distribution by Poisson. The parameters for the binomial distribution are n = and p = P (straight flush out of 5 given cards) = 4 ( 9 52 ) = [The denominator should be clear, for the numerator note that a straight flush is determined by the suit and lowest card (can be 2,... ). Thus, 4 9 possibilities to get a straight flush. Note: Also correct 4, if the ace is considered both high and low card.] Thus, P (B(, p) ) = P (P oisson(λ) ) = P (P oisson(λ) = ) = e λ, where λ = n p =.3 4
5 5 ( pts) The average height of a 25yearold man is inches. For about % of all 25yearold men, the height is within 2 inches of the mean. What percentage of the 25yearold men are over 6 feet? What percentage in the 6footer club are over 6 foot 5 inches? [N.B. foot=2 inches] Answer: Normal distribution. The parameters are given µ =, σ = 2. P (N(, 2) 2) = P (N(, ).5) =.69 =.3 For the second part, we want the conditional probability P (N(, 2) N(, 2) 2) = P (N(, 2) ) P (N(, ) 3 = P (N(, 2) 2) P (N(, ).5) =.3.3.4% 5
6 6 (5 pts) Each item produced by a certain manufacturer is, independently, of acceptable quality with probability.95. Approximate the probability that at most of the next items produced are unacceptable. Answer: Binomial distribution/approximation of the binomial by the normal distribution. We want to compute P (B(,.5) ). The parameters for the normal distribution approximating the binomial are µ = E[B(n, p)] = n p = 5 σ 2 = Var[B(n, p)] = n p( p) = 4.5 Thus, σ = 4.5 = 2.8. In conclusion, the final answer is: ( P (B(,.5) ) = P (N(5, 2.8) ) = P N(, ) 5 ) = P (N(.) 2.29) = 98.9% 2.8 6
7 ( pts) A man aiming at a target receives points if his shot is within inch of the target, 5 points if it is between and 3 inches of the target, and 3 points if it is between 3 and 6 inches of the target. Find the expected number of points scored if the distance from the shot to the target is uniformly distributed between and. Answer: If the target is assumed to consist of concentric circles, then the probabilities are, 8, 2, 64 (obtained by computing the areas of the various regions) for, 5, 3, and points respectively. Thus the expectation is E[X] = = 3 Similarly, if the target is assumed linear, then the probabilities are, 2, 3, 4, etc.
8 8 (5 pts) i) In independent tosses of a coin, the coin landed heads 58 times. Is it reasonable to assume that the coin is not fair? Answer: Approximation of the normal by the distribution. Need to compute P (N(5, 5) 58) = P (N(, ) 6) very small probability (6 standard deviations!). Thus, probably not a fair coin. ii) The average lifetime of a radio is 8 years. If Jones buys a used radio, what is the probability that will be working after an additional 8 years? Answer: Exponential distribution with parameter λ = 8. The exponential distribution is forgetful. Thus, the probability to use for at least 8 more years is the same as the probability of using a new radio for at least 8 years, i.e. ( ( ) P Exp 8 = e 8) = e iii) A point is chosen at random in the segment [, ]. Find the probability that the ratio of the shorter to longer segment is at least 2. Answer: Uniform distribution. If x denotes the distance to the left end point of the interval. We get 2 x x 2 Thus, we should have [ x 3, 2 3] The probability is the length of this segment, i.e. 3 8
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