<p> Chapter 5 Review (w/key)</p><p>1) a. With four flips of a balanced coin, there are ____ likely possibilities. Name them all below:</p><p>TTTT TTTH TTHT THTT HTTT HHTT HTHT HTTH THTH TTHH THHT HHHT HHTH HTHH THHH HHHH</p><p> b. Is this a probability distribution? Why? Yes; All probabilities are between 0 and 1, and the probabilities of the sample space adds up to 1.</p><p> c. Fill in the columns with the appropriate headings. # of Heads=L1 Probability=L2 L3=L1*L2 L1-squared L5=L4*L2 0 1/16 0 0 0 1 4/16 .25 1 .25 2 6/16 .75 4 1.5 3 4/16 .75 9 2.25 4 1/ .25 16 1</p><p> d. Find the mean. Sum L3 = 2 e. Find the variance. 5-(2^2)=1 f. Find the standard deviation. Square root of e = 1. g. You are going to flip the coin four times, and want to know the probability of getting two heads. Is this a binomial distribution? Why or why not? Yes, because there are 2 outcomes (heads/tails), a set number of trials (4), Flips are independent, and the probability of getting a head is set for every trial at .5.</p><p>2) Check whether the following probability distribution given by x + 3 P( x ) = for x = 1, 2, 3 15 Is a probability distribution of some random variable.</p><p>X = L1 Probability=L2 L3=L1*L2 L1-squared L5=L4*L2 1 4/15 2 5/15 3 3/15 No, if you sum up the probabilities you get 12/15 which is not equal to 1; therefore, not a probability distribution.</p><p>3) If the probability is 0.70 that any one registered voter (randomly selected form official roles) will vote in a given election, what is the probability that two of five registered voters will vote in the election?</p><p>Find P(X=2): Check 2NIP; its binomial. Use the binomial formula to prove that the probability is .1323 (binompdf(5, .7, 2)).</p><p>4) The probability is 0.30 that a person shopping at a certain supermarket will take advantage of its special promotion of ice cream.</p><p> a. Find the probabilities that among six persons shopping at this market there will be 0, 1, 2, 3, 4, 5, or 6 who will take advantage of this promotion. X=L1 Probability=L2 L3=L1*L2 L1-squared L5=L4*L2 0 .1176 1 .3025 2 .3241 3 .18522 4 .0595 5 .0102 6 .0007</p><p>Binompdf(6,.3,0)=.1176, Binompdf(6,.3,1)=.3025, etc….</p><p> b. Find the mean number of persons, among six shopping at the market, who will take advantage of the special promotion. Mean = np = 6 (.3) = 1.8.</p><p>5) If the probability is 0.05 that there will be a serious accident at a certain intersection on a weekday, what is the probability that there will be a serious accident at that intersection on a least three of 20 workdays?</p><p>P(X is greater than or equal to 3) = 1-binomialcdf(20, .05, 2)=.0755</p><p>6) It is known that 2 percent of the books bound at a certain bindery have defective bindings. Use the Poisson approximation to find the probability that 5 of 400 books bound by this bindery will have defective bindings.</p><p>Mean = 5/400 = .008. Using the Poisson probability formula, P(x=5) = 2.71 x 10^-13 Check with calculator: poissonpdf(.008, 5) 7) Records show that the probability is 0.00006 that a car tire will go flat while being driven through a certain tunnel. Use the Poisson approximation to find the probability that at least 2 of 10,000 cars passing through that tunnel will get flat tires.</p><p>First, calculate the mean: .00006*10,000 = .6 P(X is greater than or equal to 2) = 1-P(X is less than or equal to 1) = 1- Poissoncdf(.6, 1) = .1219</p><p>8) If the probabilities are 0.06, 0.21, 0.24, 0.18, 0.14, 0.10, 0.04, 0.02, 0.01 than at airline office at a certain airport will receive 0, 1, 2, 3, 4, 5, 6, 7, or 8 complaints per day about its luggage handling,</p><p> a. How many such complaints can it expect per day? X 0 1 2 3 4 5 6 7 8 P(x) .06 .21 .24 .18 .14 .1 .04 .02 .01 Mean of a distribution = expected value = use formula 5.1. E(X) = 2.75.</p><p> b. Find the mean of this probability distribution. Same answer as a! Expected value = mean of a probability distribution = 2.75.</p><p> c. Find the standard deviation. Use formula 5.4 to find it. Variance = 3.0275, so Standard Deviation = 1.74.</p><p>9) The probabilities are 0.22, 0.36, 0.28, and 0.14 that an investor will be able to sell a piece of property at a profit of $2,500, at a profit of $1,500, at a profit of $500, or at a loss of $500. What is the investor’s expected profit? X 2500 1500 500 -500 P(X) .22 .36 .28 .14 E(X) = Mean of a probability distribution = $1160.</p>
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