Chapter 8 Solutions Page 1 of 15 CHAPTER 8 EXERCISE SOLUTIONS

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1 Chapter 8 Solutions Page of 5 8. a. Continuous. b. Discrete. c. Continuous. d. Discrete. e. Discrete. 8. a. Discrete. b. Continuous. c. Discrete. d. Discrete. CHAPTER 8 EXERCISE SOLUTIONS 8.3 a. 3/6 = / makes the probabilities sum to. b.. makes the probabilities sum to. Calculation is ( ) = ( ) =.0. (The sum of all probabilities must equal.) 8.5 a. Condition : Sum = =. Condition : Each of the four probabilities is between 0 and. b. Figure for Exercise 8.5c c. k 0 3 P(X k)...5 As an example, P(X ) = P(X=0) + P(X=) =.+.= k P(X=k) /8 3/8 3/8 /8 8.7 a..80. Find this by adding the probabilities for X = 0,, and. P(X ) = P(X =0) + P(X=) + P(X=) = =.80. b. P(X = or X = ) = =.66. c. P(X > 0) = P(X = 0) =.4 =.86.

2 Chapter 8 Solutions Page of The probability that X=0 are left-handed is (7/0)(6/9) = 4/90 (see Section 7.4, Rule 3). The probability that X= are left-handed is (3/0)(/9) = 6/90. P(X=) must be such that probabilities add to, so it is (4/90) (6/90) = 4/90. A summary of the distribution (pdf) is: k 0 P(X=k) 4/90 4/90 6/ The answer is 8/=/. Add the probabilities given in Example 8.5 for X =, 4, 6, 8, 0, and. 8.0 The sample space of all possible sequences is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} a. For each sequence in the sample space, count the number of tails. Then count how many outcomes there are for each possible number of tails. Divide by 8 because all outcomes in the sample space are equally likely. The distribution (pdf) is: k 0 3 P(X =k) /8 3/8 3/8 /8 b. For each sequence in the sample space, take the difference (Y) between the number of heads and the number of tails. For instance, for HHT, Y = (heads) (tails) =. List all possible differences, count how many sequences in the sample space have each possible value, then divide by 8 to get the probability of that value. The distribution (pdf) is: k 3 3 P(Y =k) /8 3/8 3/8 /8 c. The sum of heads and tails is 3 for every sequence which means P(Z=3) = so the distribution (pdf) is: k 3 P(Z =k) 8. The probability that X= is the probability that the first child is a girl, so P(X=)=.5. For X=, the sequence must be Boy, Girl so P(X=)= (.5)(.5)=.5. For X=3, the sequence is Boy, Boy, Girl and this probability is (.5)(.5)(.5)=.5. The probability that X=4 can be found be subtracting the sum of the other probabilities from, and this gives P(X=4)=.5. A summary of the distribution is: k 3 4 P(X=k) This will differ for each student. An example of another continuous random variable is the amount of rainfall (in inches) during the event time. Rainfall can be any number between 0 and some maximum value. One example of another discrete random variable is the number of police, ambulance, and fire vehicles driving by with sirens. The number of vehicles can be 0,,,, and so on up to the logical maximum for the situation. 8.3 k P(X k)

3 Chapter 8 Solutions Page 3 of This may differ for each student and the explanation should support the response. Following are some suggestions. a. The cdf would be of more interest, since she would want to know if the number (and percent) of constituents who oppose the new law is less than or equal to some standard (like 50%). b. The pdf may be of more interest because it reveals what proportion of people become infected on the first exposure, what proportion of people become infected on the second exposure, etc. The cdf may be of interest to someone who has had k exposures and wants to know the probability of infection so far. c. The cdf would be of most interest, because the company would want to know things such as the probability that the drug will lower the blood pressure of less (or more) than half of the sample. d. The cdf would have more interest if the caterer wants to have a very high probability of having enough food for all guests. The caterer would probably want to choose k to be large enough that P(X>k) would be very small. A caterer who wishes to make very exact plans might be interested in the pdf, which would give a probability for each possible number, but it's more likely that the cdf would be of interest. 8.5 a. $30, the cost of the insurance. b. X = $0 with probability.90 (the probability he doesn t need to be towed during a year), and X = $00 with probability.0 (the probability he needs to be towed during a year). c. If he buys insurance, E(X) = $30. If he does not buy insurance, E(X) = $ $00.0 = $0. In the long run, he s better off to not buy the insurance. The average cost per year, without insurance will be $0, which is less than the $30 per year cost of the insurance. 8.6 a. E(X) = (0.05) + (.60) + (.30) + (3.05) =.35. b. The average number of children per family over many families similar to the Brann family. c. No. The expected value of.35 is an impossible number of children for one family. 8.7 E(X) = ($00)(.0) + ( $5) (.99) = $ a. E(X) = (.5) + 0(.50) + (.5) = 0. b. σ = ( 0) (.5) + (0 0) (.50) + ( 0) (.5) =.50. c = a. E(X) = (.5) + 0(.50) + (.5) = 0. b. σ = ( 0) (.5) + (0 0) (.50) + ( 0) (.5) =. c. σ = = An intuitive solution (and a correct one) is that the expected value for the number of girls among three children is 3.5 =.5. A more elaborate solution is to determine the probability distribution for X =number of girls, and then use the formula for E(X). The distribution (given in Example 8.4) is: k 0 3 P(X=k) /8=.5 3/8 =.375 3/8=.375 /8=.5 The expected value is calculated as the sum of "value probability" and the calculations here are, E(X) =(0.5) +(.375)+(.375) + (3.5) =.5 girls. 8. The distribution of the sum is given in Example 8.5. The expected value is µ = E ( X ) = = = 7

4 Chapter 8 Solutions Page 4 of 5 8. σ =. 45. The formula is σ = ( x µ ) p( x) where the sum is over all values of x. Here, the mean is µ = 7 (see Exercise 8.0). Details of the calculations are: x x µ (x µ) p(x) / / 3/ 4/ 5/ 6/ 5/ 4/ 3/ / / σ = = 5 0/ + 6 = σ = $ The standard deviation is calculated as σ = ( x µ ) p( x) where the sum is over all values of x. The expected value (mean) was found in Exercise 8.6 as µ = $0.35. The details for finding σ are: x $4,999 $49 $4 $0 $ x µ $4, $49.35 $4.35 $0.35 $0.65 p(x) ( ) + $49.35 (.0068 ) + $4.35 (.0303 `) + $0.35 (.4 ) + $0.65 (.76 ) = $ µ = E(X) = xp (x) = (500.) + (0.9) = $50 per person (in the long run). 8.5 Answer = $60. In Exercise 8.3, the result is that the company will pay out $50 per person in the long run so they should charge each person $50 + $0 = $60 to make a $0 per person profit. 8.6 µ = E(X) = xp (x) = (0.03) + (.5) + (.7) =.69. (The probability for living with one parent is the sum of probabilities for mother only and father only.) Here, the expected value is the average number of parents in the household over the population of all children. This value is not an "expected" value for an individual child because it is not possible to live with.69 parents. It is meaningful only as a long run or population average. 8.7 µ = E(X) = xp (x) = (3.7) + (4.3) = 3.3. You would not expect to have this grade point average every quarter or semester. Instead, it is your expected grade point average in the long run. 8.8 a. µ = E(X) = xp (x) = (5.8) + (0.) = 6 minutes. b. No, the expected value will never be your actual commute time (which is always either 5 min. or 0 min.) 8.9 a. Yes. n = 00 and p =.5. b. E(X) = np = (00)(.5) = a. Yes. n = 0 and p =.5. b. No. p is not the same from trial to trial. c. No. The trials (cities) are not independent of each other as they will tend to have the same weather. d. No. The trials (children) are not independent of each other because they are in the same class and flu is contagious.

5 Chapter 8 Solutions Page 5 of a. n = 30 and p = /6. b. n = 0 and p = /00. c. n = 0 and p = 3/ a. µ = E(X) = np = (30)(/6) = 5. b. µ = E(X) = np = (0)(/00) = 0.0. c. µ = E(X) = np = (0)(3/0) = The answers can be found using any of the methods discussed in Section 8.4, including the use of Minitab or Excel. a. P(X = 5) =.064. b. P(X = ) =.300. c. P(X = ) =.684. d. P(X = 9) = a. µ = 0(.50) = 5; σ = 0 (.5)(.5) =.58. b. µ = (.40) =0.40; σ = (.40)(.40) = c. µ = 00(.90) = 90; σ = 00 (.9)(.9) = 3. d. µ = 30(.0) = 0.3; σ = 30 (.0)(.0) = a. The probability of success does not remain the same from one trial (game) to the next. The probability of winning a game against a good team is not the same as the probability of winning a game against a poor team. b. The number of trials is not specified in advance. c. The probability of success does not remain the same from one trial to the next because whether or not the first card is an ace affects the probability that the next card is an ace, and so on. This also means that trials are not independent. 8. Use a binomial distribution with n = 0, p =.5. Let X = number of games won by human. The answers can be found using any of the methods discussed in Section 8.4, including the use of Minitab or Excel. a. P(X = 5) =.46. b. P(X = 3) =.7. If computer wins 7 games, then human wins 3 games. c. P(X 7) =.79, calculated as P(X = 7) + P(X = 8) + P(X = 9) + P(X = 0) = =.79. Equivalently, P(X 7) = P(X 6) = a. The number of trials is specified in advance. There are two possible outcomes either the subject guesses correctly or not. If the subject merely guesses, the probability of success remains the same from trial to trial. Whether a subject guesses correctly or not on a trial is independent from the results of previous trials. b. Yes, X is a binomial random variable with n = 0 and p =.5. c. The number correct is either 6 or more or 5 or less, so P(X 6) = P(X 5) =.9803 =.097. d. With p =.5, P(X 6) = P(X 5) =.630 = e. This answer will differ for each student. A factor to consider is that among all people who merely guess,.097(about %) will be able to get 6 or more correct. If many people are tested, a few who just guess will be able to get 6 or more right. Another factor to consider is the possible proportion in the population tested that actually has psychic ability. If few people have psychic ability, a result of 6 or more correct might reasonably be considered to have been the result of lucky guessing. If many people actually have psychic ability, it might be reasonable to think the result was obtained from one of those with psychic ability. The discussion of the confusion of the inverse in Section 7.7 is relevant to this discussion.

6 Chapter 8 Solutions Page 6 of The answers for this exercise can be found using any of the methods discussed in Section 8.4, including the use of Minitab or Excel. a. P(X = 4) =.05 b. P(X 4) = P(X 3) =.6496 =.3504 c. P(X 3) =.6496 d. P(X = 0) =.5905 e. P(X ) = P(X = 0) =.5905 = a. Note that /4 of,000 is 50 so the desired probability is P(X 50). n = 000 and p = the proportion of adults in the United States living with a partner, but not married at the time of the sampling. The value of p is not known. b. The desired probability is P(X 0), n = 500, and p =.0. c. Note that 70% of 0 is 4 so the desired probability is P(X 4). n = 0, and p = The formulas are µ =np and σ = np( p) a. µ = 0(/) = 5 and σ = 0 (.5)(.5) =. 58 b. µ = 00(/4) = 5 and σ = 00 (.5)(.5) = c. µ = 500(/5) = 500 and σ = 500 (.)(.) = 0 d. µ = (/0) =. and σ = (.)(.) =. 3 e. µ = 30(.4) = and σ = 30 (.4)(.4) = For n = and p =.5, P(X = 0) =.5, P(X = ) =.5, and P(X = ) =.5. These can be found in several ways. One way is to list possible outcomes, which are {SS, SF, FS, and FF}, recognize that all outcomes are equally likely, and then tabulate the distribution of X = number of successes. µ = E(X) = xp (x) = (0.5) + (.5)+ (0.5) =. σ = ( x µ ) p( x) = (0 ) (.5) + ( ) (.5) + ( ) (.5 ) = 0.5 = a. P(0 X 30) =.5 (because the interval from 0 to 30 is one-half of the interval of possible outcomes (0 to 60) and the distribution is uniform. b. P(30 X 60) =.5 by the same reasoning as in part (a) a. =.5. b. 4 0 =. 6 c. 0 0 =. 5 d. 5 ( 0) = Table A. can be used to find the answers. a b..3. c..68. d e f g