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251y0831 4/18/08 ECO251 QBA1 THIRD EXAM Apr 24, 2008 Name: ________KEY_________ Student Number: _____________________ Class time: _____________________ Part I. (16 points) Do all the following (2 points each unless noted otherwise). Make Diagrams! Show your work! In particular you must briefly explain how you got the answer to the value of z at the bottom of this page. z has the standardized Normal distribution z ~ N 0, 1 for the first four problems. 1. Pz 1.23 P1.23 z 0 Pz 0 .3907 .5 .8907 For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the entire area above -1.23. Because this is on both sides of zero, we must add the area between 0.75 and zero to the area above zero. 2. P3.25 z 0 .4994 For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade only the area between -3.25 and zero. Because this is on one side of zero and ends at zero, we can simply look up 3.25 on the table. The area between -3.25 and zero is exactly the same as the area between zero and 3.25 if the mean is zero. The bottom of our Normal table has the following. For values above 3.09, see below If z 0 is between 3.08 3.11 3.14 3.18 3.22 3.27 3.33 3.39 3.49 3.62 3.90 and and and and and and and and and and and 3.10 3.13 3.17 3.21 3.26 3.32 3.38 3.48 3.61 3.89 up P0 z z0 is .4990 .4991 .4992 .4993 .4994 .4995 .4996 .4997 .4998 .4999 .5000 Since 3.21 is between 3.22 and 3.25 we use .4994. 3. P3.07 z 3.07 P 3.07 z 0 P0 z 3.07 .4989 .4989 .9978 For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the area between -3.07 and 3.07. Because this is on both sides of zero, we must add the area between -3.07 and zero to the area between zero and 3.07. These areas are identical. 4. z .135 Solution: z .135 is, by definition, the value of z with a probability of 13.5% above it. Make a diagram. The diagram for z will show an area with a probability of 100 13 .5% 86.5% below z .135 . It is split by a vertical line at zero into two areas. The lower one has a probability of 50% and the upper one a probability of 86.5% - 50% = 36.5%. The upper tail of the distribution above z .135 has a probability of 13.5%, so that the entire area above 0 adds to 36.5% + 13.5% = 50%. From the diagram, we want one point z .135 so that Pz z .135 .86 .5 or P0 z z.135 .3650 . If we try to find this point on the Normal table, the closest we can come is P0 z 1.10 .3646 , though P0 z 1.11 .3665 is more or less acceptable. . So we will use z.135 1.10 , though 1.11 might be acceptable. 1 251y0831 4/18/08 x ~ N 4, 7 for problems 5 through 8. Note that all values of z are rounded to the nearest hundredth. On this page we must make the simple transformation z x to ‘standardize’ our values of x. 1.23 4 5. Px 1.23 P z Pz 0.75 P0.75 z 0 Pz 0 .2734 .5 .7734 7 For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the entire area above -0.75. Because this is on both sides of zero, we must add the area between -0.75 and zero to the area above zero. If you wish, make a completely separate diagram for x . Draw a Normal curve with a mean at 4. Indicate the mean by a vertical line! Shade the area above -1.23. This area is on both sides of the mean (4), so we add the area between -1.23 and the mean to the whole area (.5) above the mean . 0 4 3.25 4 z P 1.04 z 0.57 6. P3.25 x 0 P 7 7 P1.04 z 0 P0.57 z 0 .3508 .2157 .1351 For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the area between -1.04 and -0.57. Because this is entirely on one side of zero, we must subtract the smaller area between -0.57 and zero from the larger area between -1.04 and zero. If you wish, make a completely separate diagram for x . Draw a Normal curve with a mean at 4. Indicate the mean by a vertical line! Shade the area between -3.25 and 0. Both these numbers are to the left of the mean. This area is on one side of the mean (4), so we subtract the smaller area between -3.25 and the mean from the larger area between 0 and the mean. 3.07 4 3.07 4 z P 1.01 z 0.13 7. P3.07 x 3.07 P 7 7 P1.01 z 0 P0.13 z 0 .3438 .0517 .2921 For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the area between -1.01 and -0.13. Because this is entirely on one side of zero, we must subtract the smaller area between -0.13 and zero from the larger area between -1.01 and zero. If you wish, make a completely separate diagram for x . Draw a Normal curve with a mean at 4. Indicate the mean by a vertical line! Shade the area between -3.07 and 3.07. Both these numbers are to the left of the mean. This area is on one side of the mean (4), so we subtract the smaller area between 3.07 and the mean from the larger area between -3.07 and the mean. 8. x.135 Solution: x.135 is, by definition, the value of x with a probability of 13.5% above it. If you did not do so on the last page, make a diagram. The diagram for z will show an area with a probability of 100 13.5% 86.5% below z.135 . It is split by a vertical line at zero into two areas. The lower one has a probability of 50% and the upper one a probability of 86.5% - 50% = 36.5%. The upper tail of the distribution above z .135 has a probability of 13.5%, so that the entire area above 0 adds to 36.5% + 13.5% = 50%. From the diagram, we want one point z .135 so that Pz z .135 .86 .5 or P0 z z.135 .3650 . If we try to find this point on the Normal table, the closest we can come is P0 z 1.10 .3646 , though P0 z 1.11 .3665 is more or less acceptable. . So we will use z.135 1.10 , though 1.11 might be acceptable. You have gotten to this point with something like z.135 1.10 . We now must get back from z to x. The opposite of z x is x z . Since x ~ N 4, 7 , we have x.135 4 1.107 11 .70 . Check: 11 .70 4 Px 11 .70 P z Pz 1.10 Pz 0 P0 z 1.10 .5 .3643 .1357 13 .5% 7 2 251y0831 4/18/08 Part II: (9+ points) Do all the following: All questions are 2 points each except as marked. Exam is normed on 50 points including take-home. (Showing your work can give partial credit on some problems! In open-ended questions it is expected. Please indicate clearly what sections of the problem you are answering and what formulas you are using. Neatness counts!) Remember that you may not be able to finish this section, so ration your time on each problem. [Numbers in brackets are a cumulative total] 1. A small life insurance company receives an average of five death claims a day. Assume that the Poisson distribution is correct. What is the probability that the company will receive more than 10 claims in a given day (rounded to thousandths)? (2) a) .986 b) .005 c)* .014 d) .032 e) None of the above (Fill in an answer!) Explanation: From the Poisson (5) table, Px 10 .98630 . So Px 10 1 Px 10 1 .98630 = .0137 2. A local family planning group serves 20000 teen-age girls. It costs $50 to council each pregnant girl. There is a 5% chance that each of the girls will become pregnant during the year. Each pregnancy can be assumed an independent event. Based on what you know about the expected values of discrete distributions, how much should the agency budget for counseling this year? (2) a) $1000 b) $20000 c) *$50000 d) $100000 e) None of the above. Explanation: The Binomial distribution applies since we have independent events with a constant probability. n 20000 , p .05 . If x is the number of girls pregnant, E x np .0520000 1000 . The cost, however, is C 50 x , so E50 x 50 E x 501000 50000 . 3. In the agency in problem two, 20 girls are waiting to see a counselor this morning. Half of them are pregnant. If Samantha is assigned to counsel 8 of them, what is the chance that all eight are pregnant? (I want to see your formulas and calculations – this only took me a few minutes.) (3) Explanation: This is a hypergeometric problem. N 20 , M Np 10 and n 8. 10 9 10! 1 C 810 C 010 2!8! 21 P8 .00036 . If you need more of an explanation: there are 20 20 ! 20 19 18 17 16 1514 13 C8 12!8! 87654321 10 9 10 ! C810 45 ways to pick 8 pregnant girls from the 10 that are pregnant and 2!8! 21 C 820 10! 125970 ways to pick 8 girls from the 20 waiting. 12!8! 4. How many girls would have to be waiting before we could use the Binomial distribution to solve problem 3? (1) [8] Solution: There would have to be more than N 20 n 208 160 3 251y0831 4/18/08 5. OK. Assume that there are 4000 girls waiting to see a counselor and Samantha is assigned to counsel 8 of them, what is the chance that all eight are pregnant? The only answer that I will accept here is an answer gotten from numbers in the tables. If you just write down a solution, you will get half credit. (2) Solution: Use the binomial table with p .5 and n 8 . Px 8 Px 8 Px 7 1 .99609 .00391 6. A variable has the Binomial distribution . Find the following and explain how you did it. a) Px 5 p .02 n 200 (2) n 200 Solution: Use the Poisson distribution because 10000 500 . m np 4 . p .02 Px 5 1 Px 4 1 .62884 .37116 b) Px 60 p .40 n 200 (2 or 2.5) [14] Solution: Use Normal distribution because np 200 .40 80 5 and nq 200 .60 200 80 120 5 . 2 npq 80.6 48 . Without continuity 60 80 correction Px 60 P z Pz 2.89 .5 .4981 .9981 . With continuity correction 48 59 .5 80 Px 60 P z Pz 2.96 .5 .4985 .9945 48 7. Find P11 x 17 for the following distribution: Continuous Uniform with c 12 , d 15 (1) Solution: c 12 and d 15 . Make a diagram. The base of the box goes from c 12 to d 15 . Shade the area inside the box between 11 and 17. 0 11 12 15 17 1 1 1 The height of the box is and the base of the shaded area is 15 – 12 = 3. So the area of d c 15 12 3 3 the box is P11 x 17 = 1. 3 xc for c x d and If you use the cumulative distribution method, remember F x 1 for x d , F x d c F x 0 for x d . So P11 x 17 F 17 F 11 1 0 1 . Never forget that all probabilities are between 0 and 1. If you get anything else, you have misused a formula. 8. If the amount of time it takes students to find a parking place has a Normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, 99% of students search time will fall below what number? (For example it may be true that 99% take 3.6 or fewer minutes to find a parking place, but I doubt it.) (2) Solution: According to the t-table z.01 2.327 . So x.01 z.01 3.5 2.327 1 5.827 . 4 251y0831 4/18/08 9. (Dummeldinger) You decide to invest in four independently moving risky stocks. You guess that each stock has an independent 40% chance of becoming a total loss. What is the chance that at least one of your stocks will tank? (2) [19] a) 0 a) .0256 b) .4000 c)* .8704 d) .9744 e) 1.0000 Solution: c. Binomial p .40 n 4. 1 P0 1 .1296 .8704 10. (Extra Credit) The amount of rainfall in a 24 – hour period has an exponential distribution with a mean of 0.2 inches. What is the chance that a randomly picked 24-hour period will have rainfall that exceeds 0.8 inches? a)* .018316 b) .778801 c) .221199 d) .981684 e) None of the above. Produce an alternate answer. Solution: 1 Px 0.8 1 (1 e 5.8 ) .018316 Blank page for calculations. 5 251y0831 4/18/08 ECO251 QBA1 THIRD EXAM Apr 24, 2008 TAKE HOME SECTION Name: _________________________ Student Number: _________________________ Throughout this exam show your work! Please indicate clearly what sections of the problem you are answering and what formulas you are using. Write on one side of the page! Part III. Do all the Following (25+ Points) Show your work! Neatness counts! Answers of ‘zero’ or ‘one’ especially are unacceptable without an explanation. Do not use one distribution to approximate another without justifying the replacement! 1. Identify the distribution that you are using in each problem. If you have a number like n 20 g , make it very clear what value of n you are using. Look at the solved problems for Section L, the solution to Grass3 and ‘Great Distributions’ (especially the hints on the 3 rd page) before you start. Let g be the last digit of your student number. If g 0 , change it to 2. Hint: It may help in these problems to realize that if n 100 , 93 or more is 93 to 100 and 10 or less is 10 to zero. This is especially useful if we count failures. a. A basketball player makes 50 5g % of her free throws over the basketball season. In one game she gets 20 free throws and misses 13 g of them. The coach will investigate if the probability of doing as badly as she did or worse is below 5%. Will the coach investigate? Do the math to justify your answer. (2) Solution: This is a hypothesis test. We are testing the null hypothesis H 0 : p .50 .05 g against the alternative H 1 : p .50 .05 g . An ‘s’ indicates that p the observed proportion (8/20 = .40 in the first case) is significantly below p 0 .50 .05 g , so that the coach will investigate. An ‘ns’ indicates that p is not significantly below p 0 . Note that, though in the majority of cases we cannot reject the null hypothesis, if we had the same proportion of a larger sample we could reject it. g 1. She misses 12. Her probability of hitting is 55%. Her probability of missing is 100 – 55 = 45% Px 12 n 20; p .45 1 Px 11 1 .86923 = .1308 ns g 2. She misses 11. Her probability of hitting is 60%. Her probability of missing is 100 – 60 = 40% Px 11 n 20; p .40 1 Px 10 1 .87248 = .1275 ns g 3. She misses 10. Her probability of hitting is 65%. Her probability of missing is 100 – 65 = 35% Px 10 n 20; p .35 1 Px 9 1 .87822 = .1218 ns g 4. She misses 9. Her probability of hitting is 70%. Her probability of missing is 100 – 70 = 30% ns Px 9 n 20; p .30 1 Px 8 1 .88667 = .1122 g 5. She misses 8. Her probability of hitting is 75%. Her probability of missing is 100 – 75 = 25% ns Px 8 n 20; p .25 1 Px 7 1 .89819 = .1018 g 6. She misses 7. Her probability of hitting is 80%. Her probability of missing is 100 – 80= 20% ns Px 7 n 20; p .20 1 Px 6 1 .91331 = .0867 g 7. She misses 6. Her probability of hitting is 85%. Her probability of missing is 100 – 85 = 15% ns Px 6 n 20; p .15 1 Px 5 1 .93269 = .0673 g 8. She misses 5. Her probability of hitting is 90%. Her probability of missing is 100 – 90 = 10% Px 5 n 20; p .10 1 Px 4 1 .95683 = .0432 s g 9 . She misses 4. Her probability of hitting is 95%. Her probability of missing is 100 – 95 = 5%. Px 4 n 20; p .05 1 Px 3 1 .98412 = .0159 s 6 251y0831 4/18/08 b. We believe that 80% of all accidents involve alcohol. If we investigate 6 g accidents, what is the probability that 2 g or fewer involve alcohol? (1) Solution: This can also be interpreted as a hypothesis test. We are testing the null hypothesis H 0 : p .80 against the alternative H 1 : p .80 . An ‘s’ indicates that p the observed proportion (3/7 = .4286 in the first case) is significantly below p 0 .80 , so that we can doubt the null hypothesis. An ‘ns’ indicates that p is not significantly below p 0 . Note that, though in all cases except the first we cannot reject the null hypothesis, if we had the same proportion of a larger sample we could reject it. g 1. There are 7 accidents of which 3 or fewer involve alcohol. This is zero to 3 involving alcohol, or 4 to 7 not involving alcohol. The probability of an accident not involving alcohol is .20. ns P x 3 n 7; p .80 P x 4 n 7; p .20 1 Px 3 1 .96666 = .0333 g 2. There are 8 accidents of which 4 or fewer involve alcohol. This is zero to 4 involving alcohol, or 4 to 8 not involving alcohol. The probability of an accident not involving alcohol is .20. P x 4 n 8; p .80 P x 4 n 8; p .20 1 Px 3 1 .94374 = .0563 ns g 3. There are 9 accidents of which 5 or fewer involve alcohol. This is zero to 5 involving alcohol, or 4 to 9 not involving alcohol. The probability of an accident not involving alcohol is .20. P x 5 n 9; p .80 P x 4 n 9; p .20 1 Px 3 1 .91436 = .0859 ns g 4. There are 10 accidents of which 6 or fewer involve alcohol. This is zero to 6 involving alcohol, or 4 to 10 not involving alcohol. The probability of an accident not involving alcohol is .20. P x 6 n 10; p .80 P x 4 n 10; p .20 1 Px 3 1 .87913 = .1206 ns g 5. There are 11 accidents of which 7 or fewer involve alcohol. This is zero to 7 involving alcohol, or 4 to 11 not involving alcohol. The probability of an accident not involving alcohol is .20. P x 7 n 11; p .80 P x 4 n 11; p .20 1 Px 3 1 .83886 = .1611 ns g 6. There are 12 accidents of which 8 or fewer involve alcohol. This is zero to 8 involving alcohol, or 4 to 12 not involving alcohol. The probability of an accident not involving alcohol is .20. P x 8 n 12; p .80 P x 4 n 12; p .20 1 Px 3 1 .79457 = .2054 ns g 7. There are 13 accidents of which 9 or fewer involve alcohol. This is zero to 9 involving alcohol, or 4 to 13 not involving alcohol. The probability of an accident not involving alcohol is .20. P x 9 n 13; p .80 P x 4 n 13; p .20 1 Px 3 1 .74732 = .2527 ns g 8. There are 14 accidents of which 10 or fewer involve alcohol. This is zero to 10 involving alcohol, or 4 to 14 not involving alcohol. The probability of an accident not involving alcohol is .20. ns P x 10 n 14; p .80 P x 4 n 14; p .20 1 Px 3 1 .69819 = .3018 g 9. There are 15 accidents of which 11 or fewer involve alcohol. This is zero to 11 involving alcohol, or 4 to 15 not involving alcohol. The probability of an accident not involving alcohol is .20. ns P x 11 n 15; p .80 P x 4 n 15; p .20 1 Px 3 1 .64816 = .3518 7 251y0831 4/18/08 c. A test consists of 6 g questions and a student must get at least 70% of the questions right to pass the course. Each question is a multiple choice question with 2 possible answers. Note that if n 16 , 70% of 16 is 11.2, so 12 or more must be right. What is the probability of passing the exam if you just guess? What happens to the probability as the size of the exam increases to 20 questions? (2) Assume that the professor instead offers an exam with 6 g questions with four choices and then says that a passing grade will be 50% or more. Does this raise or lower the chance of passing for the guesser? What happens to this probability as the size of the exam rises to 20? (2) Solution: g 1, n 7, p .5 . Passing is 5 or more. P x 5 n 7, p .5 1 Px 4 1 .77344 .2266 p .25 . Passing is 4 or more. Px 4 n 7, p .25 1 Px 3 1 .92944 .0706 g 2, n 8, p .5 . Passing is 6 or more. P x 6 n 8, p .5 1 Px 5 1 .85547 .1445 p .25 . Passing is 4 or more. P x 4 n 8, p .25 1 Px 3 1 .88618 .1138 g 3, n 9, p .5 . Passing is 7 or more. P x 7 n 9, p .5 1 Px 6 1 .91016 .0898 p .25 . Passing is 5 or more. P x 5 n 9, p .25 1 Px 4 1 .95107 .0489 g 4, n 10, p .5 . Passing is 7 or more. P x 7 n 10, p .5 1 Px 6 1 .82810 .1719 p .25 . Passing is 5 or more. P x 4 n 10, p .25 1 Px 4 1 .92187 .0781 g 5, n 11, p .5 . Passing is 8 or more. P x 8 n 11, p .5 1 Px 7 1 .88672 .1133 p .25 . Passing is 6 or more. P x 6 n 11, p .25 1 Px 5 1 .96567 .0343 g 6, n 12, p .5 . Passing is 9 or more. P x 9 n 12, p .5 1 Px 8 1 .92700 .0730 p .25 . Passing is 6 or more. P x 6 n 12, p .25 1 Px 5 1 .94560 .0544 g 7, n 13, p .5 . Passing is 10 or more. P x 10 n 13, p .5 1 Px 9 1 .95386 .0461 p .25 . Passing is 7 or more. P x 7 n 13, p .25 1 Px 6 1 .97571 .0243 g 8, n 14, p .5 . Passing is 10 or more. P x 10 n 14, p .5 1 Px 9 1 .91022 .0898 p .25 . Passing is 7 or more. P x 7 n 14, p .25 1 Px 6 1 .96173 .0383 g 9, n 15 , p .5 . Passing is 11 or more. P x 11 n 15, p .5 1 Px 10 1 .94077 .0592 p .25 . Passing is 8 or more. P x 8 n 15, p .25 1 Px 7 1 .98270 .0173 p .25 . Passing is 10 or more. Px 10 n 10, p .25 1 Px 9 1 .98614 .0139 If n 20, p .5 . Passing is 14 or more. P x 14 n 10, p .5 1 Px 13 1 .94234 .0577 Though the path is not even the probability of passing seems to be falling as the number of questions rises. If we try to use a Normal approximation, we see that we have the probability that z is greater than a positive number that rises as n rises, but much more slowly. With both of these, when the sample size rises above 100, the probability will be smaller than Pz 4 , and this probability is less than .0001. .7 .5 .2 n P p .7 p .5 P z P z P z .4000 n .5 .5.5 n .5 .25 .25 P p .5 p .25 P z P z P z .5774 n .4330 .25 .75 n n 8 251y0831 4/18/08 d. (Dummeldinger) The diameter of ball bearings has a continuous uniform distribution over the interval 3 .1g millimeters to 5 .1g millimeters. (i) If ball bearings are only acceptable if they are between 4 and 5 millimeters, what proportion of those produced is defective? So c 3 .1g and d 5 .1g . Make a diagram. The base of the box goes from c 3 .1g to d 5 .1g . Shade the area between 4 and d 5 .1g . 0 3+.1g 4 5-.1g 5 1 1 1 The height of the box is and the base of the shaded area is d c 5 .1g 3 .1g 2 .2 g 5 .1g 4 1 .1g . So the area of the box is P4 x 5 1 .1g .5 . 2 .2 g If we use the cumulative distribution method, remember F x 1 for x d and F x 4 (3 .1g ) 1 .1g xc for c x d . So P4 x 5 F 5 F 4 1 .5 . d c 2 .2 g 2 .2 g So let g 1. Just to check my work above, make a diagram. c 3 .1g 3.1 , d 5 .1g 4.9 . 0 3.1 4 4.9 5 1 1 1 The height of the box is and the base of the shaded area is d c 4.9 3.1 1.8 1 .1g .9 .5 . Thus P4 x 5 4.9 4 0.9 . So the area of the box is P 4 x 5 1 .8 2 .2 g 1 0.1 9 .5 For g 9. c 3 .1g 3.9 and d 5 .1g 4.1 . The height of the box is 2 0.2 1.8 1 1 1 and the base of the shaded area is 4.1 4 .1 . So the area of the box is d c 4.1 3.9 0.2 .0.1 P4 x 5 .5 . 0.2 (ii) If I take a sample of three ball bearings from the assembly line. What is the chance that none of the ball bearings are defective? Solution: From the table with n 3 and p .5 P0 .1250 (iii) What is the chance that exactly one is defective? Solution: From the table with n 3 and p .5 P1 Px 1 Px 0 .5 .1250 .3750 (iv) What is the chance that at least one is defective? Solution: From the table with n 3 and p .5 Px 1 1 Px 0 1 .1250 .8750 (v) What is the chance that the third ball bearing that I inspect is the first one that is defective? Solution: From the geometric distribution formula. P3 q 2 p .53 .1250 [12] 9 251y0831 4/18/08 e. Supposedly 60% of drivers wear their seatbelts. 1000 drivers are stopped. (i) What is the approximate probability that less than half were wearing seatbelts? (1) We can approximate the binomial distribution by the Normal distribution because np 1000 (.6) 600 5 and nq 1000 600 400 5 . The standard deviation of p is pq n .6.4 .24 .00024 .015492 . So we can say that p ~ N .6, .015492 . Without a 1000 1000 .5 .6 Pz .6.46 0 . To use a continuity continuity correction P p .5 P z .01549 correction we would make the interval 0 x 499 wider by adding .5 at both ends, so that it would be 0.5 x 499 .5 . The bottom number is, frankly, silly. So we could say .4995 .6 P p .5995 P z Pz .6.49 0 .01549 (ii) What is the approximate probability that less than 200 g were wearing seatbelts? (1)[14] x ~ N np, npq , np 1000 .6 600 , npq 600 .4 240 15 .4919 200 600 For g 1 , without a continuity correction we get Px 201 Px 200 P z 15 .4919 Pz 25 .82 0 . Pretty obviously, we will get essentially the same result for any other value of g and a continuity correction won’t affect things. f. If each member of the State Highway patrol writes an average of 10 tickets a day, (i) What is the chance that a given policeman will write more than 10 g tickets in a day? (1) Use the Poisson distribution with a parameter of 10. If g 1, Px 11 1 Px 11 1 .69678 = .3032 If g 2, Px 12 1 Px 12 1 .79156 .2984 If g 3, Px 13 1 Px 13 1 .86446 .1355 If g 4, Px 14 1 Px 14 1 .91654 .0835 If g 5, Px 15 1 Px 15 1 .95126 .0487 If g 6, Px 16 1 Px 16 1 .97296 .0294 If g 7, Px 17 1 Px 17 1 98572 .0143 If g 8, Px 18 1 Px 18 1 .99281 .0072 If g 9, Px 19 1 Px 19 1 .99655 = .0035 (ii) Assuming that a policeman averages 50 tickets during a work week, what is the chance that a given policeman will write more than 510 g tickets in a week? (I strongly doubt that the answer is the same as the answer to (i)) (2) 55 .5 50 If g 1, with the continuity correction Px 55 Px 55 .5 P z 50 5.5 P z Pz 0.78 .5 .2823 .2177 . Without the continuity correction 7 . 0711 56 50 6 Px 55 Px 56 P z Pz 0.85 .5 .3023 .1977 . P z 7 . 0711 50 10 251y0831 4/18/08 60 .5 50 If g 2, with the continuity correction Px 60 Px 60 .5 P z 50 11 .5 P z Pz 1.48 .5 .4306 .0694 . Without the continuity correction 7 .0711 61 50 11 Px 60 Px 61 P z Pz 1.56 .5 .4406 .0594 . P z 7 . 0711 50 65 .5 50 If g 3, with the continuity correction Px 65 Px 65 .5 P z 50 15 .5 P z Pz 2.19 .5 .4857 .0143 . Without the continuity correction 7 .0711 66 50 16 Px 65 Px 66 P z Pz 2.26 .5 .4881 .0119 . P z 7 . 0711 50 70 .5 50 If g 4, with the continuity correction Px 70 Px 70 .5 P z 50 20 .5 P z Pz 2.90 .5 .4981 .0019 . Without the continuity correction 7.0711 71 50 21 Px 71 Px 71 P z Pz 2.96 .5 .4985 .0015 . P z 7.0711 50 75 .5 50 If g 5, with the continuity correction Px 75 Px 75 .5 P z 50 25 .5 P z Pz 3.61 .5 .4998 .0002 . Without the continuity correction 7.0711 76 50 26 Px 75 Px 76 P z Pz 3.68 .5 .4999 .0001 . P z 7.0711 50 80 .5 50 If g 6, with the continuity correction Px 80 Px 80 .5 P z 50 30 .5 P z Pz 4.31 .5 .5 0 . Without the continuity correction 7.0711 81 50 31 Px 80 Px 81 P z Pz 4.38 .5 .5 0 . P z 7 . 0711 50 85 .5 50 If g 7, with the continuity correction Px 85 Px 85 .5 P z 50 35 .5 P z Pz 5.02 .5 .5 0 . Without the continuity correction 7 .0711 81 50 31 Px 80 Px 81 P z Pz 4.38 .5 .5 0 . P z 7 . 0711 50 11 251y0831 4/18/08 90 .5 50 If g 8, with the continuity correction Px 90 Px 90 .5 P z 50 40.5 P z Pz 5.73 .5 .5 0 . Without the continuity correction 7.0711 91 50 41 Px 90 Px 91 P z Pz 5.79 .5 .5 0 . P z 7.0711 50 95 .5 50 If g 9, with the continuity correction Px 95 Px 95 .5 P z 50 45 .5 P z Pz 6.43 .5 .5 0 . Without the continuity correction 7.0711 96 50 46 Px 95 Px 96 P z Pz 6.51 .5 .5 0 . P z 7.0711 50 g. A ‘psychic’ is shown 5 cards and then is expected to be able to guess which one of the cards is randomly drawn. If the ‘psychic’ is a fake and is merely randomly guessing what card is picked and the experiment is repeated 20 times, what is the chance that the psychic will pick the right card 4 g or more times? (1) Note that this can be interpreted as a hypothesis test too. If we assume that the probability of one success is at most 20% then the probability of 8 or more successes is below 5% and we should doubt the correctness of the 5% probability. If there are 10 or more success the probability falls to 1% and we strongly doubt the null hypothesis. If g 1 , n 20 , p .2 and Px 5 1 Px 4 1 .62965 .37035 If g 2, n 20 , p .2 and Px 6 1 Px 5 1 .80421 .1958 If g 3, n 20 , p .2 and Px 7 1 Px 6 1 .91331 .0867 If g 4, n 20 , p .2 and Px 8 1 Px 7 1 .96786 .0321 If g 5, n 20 , p .2 If g 6, n 20 , p .2 If g 7, n 20 , p .2 If g 8, n 20 , p .2 and Px 9 1 Px 8 1 .99002 .0100 and Px 10 1 Px 9 1 .99741 .0026 and Px 11 1 Px 10 1 .99944 .0008 and Px 12 1 Px 11 1 .99990 .0001 If g 9 , n 20 , p .2 and Px 13 1 Px 12 1 .99998 .00002 h. The average number of small business bankruptcies in Fredonia are 13 .1g a month. What is the probability of at least one bankruptcy in a month? What is the probability of at least 3 in a month? What is the probability of more than 180 in a year? (3) This is a Poisson distribution with Px P2 e m m 2 mP 1 . 2! 2 e m m x e m m . This means P0 e m , P1 mP0 and x! 1! If g 1 , m 13.1, P0 e 13.1 .000002, P1 13.1.000002 .000027 and P2 13 .1 P1 2 .000175 So Px 1 1 P0 .999998 and Px 3 1 P0 P1 P2 .999796 181 157 .2 Pz 1.90 .5 .4713 = .0287 Px 181 m 13 .112 157 .2 P z 157 .2 12 251y0831 4/18/08 If g 2, m 13.2, P0 e 13.2 .000002 , P1 13.2.000002 .000024 and P2 13 .2 P1 2 .000161 So Px 1 1 P0 .999998 and Px 3 1 P0 P1 P2 .999806 181 158 .4 Pz 1.80 .5 .4641 = .0359 Px 181 m 13 .212 158 .4 P z 158 .4 If g 3, m 13.3, P0 e 13.3 .000002 , P1 13.3.000002 .000022 and P2 13 .3 P1 2 .000148 So Px 1 1 P0 .999998 and Px 3 1 P0 P1 P2 .999828 181 159 .6 Pz 1.69 .5 .4545 = .0455 Px 181 m 13 .312 159 .6 P z 159 .6 If g 4, m 13.4, P0 e 13.4 .000002 , P1 13.4.000002 .000020 and P2 13 .4 P1 2 .000272 So Px 1 1 P0 .999998 and Px 3 1 P0 P1 P2 .999706 181 160 .8 Pz 1.59 .5 .4441 = .0559 Px 181 m 13 .412 160 .8 P z 160 .8 If g 5, m 13.5, P0 e 13.5 .000001, P1 13.5.000002 .000019 and P2 13 .5 P1 2 .000125 So Px 1 1 P0 .999999 and Px 3 1 P0 P1 P2 .999855 181 162 .0 Pz 1.49 .5 .4319 = .0681 Px 181 m 13 .512 162 .0 P z 162 .0 If g 6, m 13.6, P0 e 13.6 .000001, P1 13.6.000001 .000017 and P2 13 .6 P1 2 .000115 So Px 1 1 P0 .999999 and Px 3 1 P0 P1 P2 .999862 181 163 .2 Pz 1.39 .5 .4177 = .0823 Px 181 m 13 .612 163 .2 P z 163 .2 If g 7, m 13.7, P0 e 13.7 .000001, P1 13.7.000001 .000015 and P2 13 .7 P1 2 .000105 So Px 1 1 P0 .999999 and Px 3 1 P0 P1 P2 .999879 181 164 .4 Pz 1.29 .5 .4015 = .0985 Px 181 m 13 .712 164 .4 P z 164 .4 If g 8, m 13.8, P0 e 13.8 .000001, P1 13.8.000001 .000014 and P2 13 .8 P1 2 .000097 So Px 1 1 P0 .999999 and Px 3 1 P0 P1 P2 .999796 181 165 .6 Pz 1.20 .5 .3849 = 1151 Px 181 m 13 .812 165 .6 P z 165 .6 If g 9 , m 13.9, P0 e 13.9 .000001, P1 13.9.000001 .000013 and P2 13 .9 P1 2 .0000898 So Px 1 1 P0 .999999 and Px 3 1 P0 P1 P2 .999796 181 166 .8 Pz 1.10 .5 .3643=.1357 Px 181 m 13 .912 166 .8 P z 166 .8 [21] 13 251y0831 4/18/08 i. There are exactly 14 firms in the US that produce jorcillators and 1 g produce computers as well. If you pick a random sample of 3 firms to examine, what is the chance that at least one will be a computer producer . What is the chance that all the firms in your sample are computer producers? Assume that the same proportion of 70 companies produce computers what would the probabilities be that at least one and all three in a 3-firm sample be computer producers? (4) [25] If g 1, 2 out of 14 produce computers and the remaining 12 do not. n 3 . So 12 1110 1 C 02 C 312 3 21 12 11 10 (i) Px 1 1 P0 1 1 1 1 .6044 .3956 (ii) This is impossible. 14 14 13 12 14 13 12 C3 3 21 2 .14286 , q 1 p .8571 . Since the sample is less than 5% of 14 the population, use the binomial distribution Px C xn p x q n x . There are only two available. (iii) p Px 1 1 P0 1 C 03 p 0 q 3 1 .8571 3 1 .6297 .3703 . (iv) P3 C 33 p 3 q 0 .14286 3 .0061 If g 2, 3 out of 14 produce computers and the remaining 11 do not. n 3 . So 11 10 9 1 3 11 C C 3 21 11 10 9 (i) Px 1 1 P0 1 0 143 1 1 1 .4533 .5467 14 13 12 14 13 12 C3 3 21 (ii) P3 C 33 C 011 C 314 3 1 .0027 (iii) p .2143 , q 1 p .7857 . Since the sample is less 14 13 12 14 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . Px 1 1 P0 1 C 03 p 0 q 3 1 .7857 3 1 .4850 .5150 . (iv) P3 C33 p 3 q 0 .2143 3 .0098 If g 3, 4 out of 14 produce computers and the remaining 10 do not. n 3 . So 10 98 1 4 10 C C 3 21 10 9 8 (i) Px 1 1 P0 1 0 143 1 1 1 .3297 .6703 14 13 12 14 13 12 C3 3 21 (ii) P3 C 34 C 010 C 314 4 4 .0109 (iii) p .2857 , q 1 p .7243 . Since the sample is less 14 13 12 14 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . Px 1 1 P0 1 C 03 p 0 q 3 1 .7243 3 1 .3800 .6200 . P3 C33 p 3 q 0 .2857 3 .0233 14 251y0831 4/18/08 If g 4, 5 out of 14 produce computers and the remaining 9 do not. n 3 . So 9 87 1 C 05 C 39 3 21 9 8 7 (i) Px 1 1 P0 1 14 1 1 1 .2308 .7692 14 13 12 14 13 12 C3 3 21 5 4 C 35 C 09 5 2 1 .3571 , q 1 p .6429 . Since the sample is less (ii) P3 .0275 (iii) p 14 14 13 12 14 C3 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . Px 1 1 P0 1 C 03 p 0 q 3 1 .6429 3 1 .2657 .7343 . (iv) P3 C33 p 3 q 0 .3571 3 .0455 If g 5, 6 out of 14 produce computers and the remaining 8 do not. n 3 . So 8 76 1 6 8 C C 3 21 87 6 (i) Px 1 1 P0 1 0 143 1 1 1 .1538 .8462 14 13 12 14 13 12 C3 3 21 65 4 C 36 C 08 6 .4286 , q 1 p .5714 . Since the sample is less (ii) P3 3 2 1 .0549 (iii) p 14 14 13 12 14 C3 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . Px 1 1 P0 1 C 03 p 0 q 3 1 .5714 3 1 .1866 .8134 . (iv) P3 C33 p 3 q 0 .4286 3 .0787 If g 6, 7 out of 14 produce computers and the remaining 7 do not. n 3 . So 7 6 5 1 7 7 C C 3 21 765 (i) Px 1 1 P0 1 0 143 1 1 1 .0962 .9038 14 13 12 14 13 12 C3 3 21 765 C 37 C 07 7 .5000 , q 1 p .5000 . Since the sample is less (ii) P3 3 2 1 .0962 . (iii) p 14 14 13 12 14 C3 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . Px 1 1 P0 1 C 03 p 0 q 3 1 .5000 3 1 .1250 .8750 . (iv) P3 C33 p 3 q 0 .5000 3 .1250 If g 7, 8 out of 14 produce computers and the remaining 6 do not. n 3 . So (i) Px 1 1 P0 1 C 08 C 36 C 314 65 4 3 21 65 4 1 1 1 .0549 .9451 14 13 12 14 13 12 3 21 1 87 6 8 .5714 , q 1 p .4286 . Since the sample is less (ii) P3 3 2 1 .1538 (iii) p 14 14 13 12 14 C3 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . C 38 C 06 Px 1 1 P0 1 C 03 p 0 q 3 1 .4286 3 1 .0787 .9213 . (iv) P3 C33 p 3 q 0 .5714 3 .1866 15 251y0831 4/18/08 If g 8, 9 out of 14 produce computers and the remaining 5 do not. n 3 . So (i) Px 1 1 P0 1 5 43 3 21 5 43 1 1 1 .0274 .9725 14 13 12 14 13 12 3 21 1 C 09 C 35 C 314 9 8 7 9 .6429 , q 1 p .3571 . Since the sample is less (ii) P3 3 2 1 .2308 . (iii) p 14 14 13 12 14 C3 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . C 39 C 05 Px 1 1 P0 1 C 03 p 0 q 3 1 .3571 3 1 .0456 .9544 . (iv) P3 C33 p 3 q 0 .6429 3 .2657 . If g 9, 10 out of 14 produce computers and the remaining 4 do not. n 3 . So (i) Px 1 1 P0 1 C 010 C 34 C 314 4 3 2 3 21 4 3 2 1 1 1 .0110 .9890 14 13 12 14 13 12 3 21 1 10 9 8 10 .7143 , q 1 p .2857 . Since the sample is less (ii) P3 3 2 1 .3297 (iii) p 14 14 13 12 14 C3 3 2 1 than 5% of the population, use the binomial distribution Px C xn p x q n x . C 310 C 04 Px 1 1 P0 1 C 03 p 0 q 3 1 .2857 3 1 .0233 .9767 . (iv) P3 C33 p 3 q 0 .7143 3 .3645 j. (Extra credit) The time between landings at an airport follows an exponential distribution with a mean of 30 seconds. What is the probability that there will be a gap of between 15 g and 45 g seconds before the next plane lands? (2) The following comes from ‘Great Distributions I have Known.’ Distribution Uses Formula Mean Variance Continuous Distributions Exponential x is usually the f x ce cx and amount of time 1 1 cx you have to wait F x 1 e c c when x 0 and until a success. the mean time to a 1 success is . Both c are zero if x 0 . 45 g 30 1 1 So 30 and c . P15 g x 45 g 1 e c 30 If g 1, P16 x 46 e 16 30 46 e 30 24 54 15 g 1 e 30 15 g e 30 45 g e 30 e 0.5333 e 1.5333 .58665 .21582 .3708 If g 9, P24 x 54 e 30 e 30 e 0.8 e 1.8 .44933 .16530 .2840 All other answers should be between these two. 16 251y0831 4/18/08 2. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the jorcillator only works as long as either component works (so that it fails in the first period only if both components fail). The probability of the phillinx failing is given by a Normal distribution with 18 g and 3. , For example, if the life of the phillinx is represented by x1 , the chance of the phillinx failing in the first ten years is P0 x1 10 and the probability of it failing in the second ten years is P10 x 20 For example: Ima Badrisk has the number 375292, so her distribution has a mean of 18 .g 20 . The flubberall has exactly the same distribution. In order to maintain my sanity, use the following events. Period 1 is the first ten years, period 2 is the second ten years and period three is happily ever after. Failure of the phillinx in period 1, 2, 3 are events A1, A2 and A3 Failure of the flubberall in period 1, 2, 3 are events B1, B 2 and B 3 Failure of the jorcillator in period 1, 2, 3 are events C1, C 2 and C 3 . a) What is the probability that the phillinx will fail in period 1? Period 2? Period 3? (2) Note that, because of the distance zero is from the mean, P0 x 10 Px 10 . I was wrong to say that zero should not be in this interval, but this did not affect any grades. 10 18 If g 0 , 18 . P A1 can be written as Px 10 P z Pz 2.67 .5 .4962 3 20 18 10 18 z = .0038. P A2 P10 x 20 P P 2.67 z 0.67 = .4962 + .2486 3 3 20 18 = .7448. P A3 Px 20 P z Pz 0.67 = .5 - .2486 = .2514. These add to 1. 3 10 27 If g 9 , 27 . P A1 can be written as Px 10 P z Pz 5.67 .5 .5 0 3 20 27 10 27 z P A2 P10 x 20 P P 5.67 z 2.33 = .5 - .4901 = .0099. 3 3 20 27 = .7448. P A3 Px 20 P z Pz 2.33 = .4901 +.5 = .9901. These add to 1. 3 For other values of g, probabilities fall between the numbers above. They always add to 1. Joint events are below. Joint event Probability if g 0 downs machine in period 1 .0038(.0038) = .00001 P A1 B1 P A1 B2 P A1 B3 P A2 B1 P A2 B2 P A2 B3 P A3 B1 P A3 B2 P A3 B3 Probability if g 9 0(0) = 0 2 .0038(.7448) = .00283 0(.0099) = 0 3 .0038(.2514) = .00096 0(.9901) = 0 2 .7448(.0038) = .00283 .0099(0) = 0 2 .7448(.7448) = .55473 .0099(.0099) = .00010 3 .7448(.2514) = .18724 .0099(.9901) = .00980 3 .2514(.0038) = .00096 .9901(0) = 0 3 .2514(.7448) = .18724 .9901(.0099) = .00980 3 .2514(.2514) = .06320 .9901(.9901) = .98030 17 251y0831 4/18/08 If we add probabilities for each period we get the table below. Probability if g 0 PC1 = .00001 PC 2 = .56039 PC3 = .43960 Probability if g 9 PC1 = 0 PC 2 = .00010 PC3 = .99990 For b), c) and d) see above b) What is the probability that the jorcillator will fail in the first period? (1) c) What is the probability that the jorcillator will fail in the second period 2? (1) d) What is the probability that the jorcillator will fail in the third period? (1) If you haven’t figured it out already, one of the easiest ways to do this is to make a joint probability table. Put the A events across the top. Put the B events down the side. Figure out what the probability of the joint events must be if they are independent. Now make a similar table. This time, instead of probabilities, fill in the period in which the jorcillator fails. e) (Extra Credit) Find the probability that the jorcillator and the Phillinx both fail in the third period (1). This is PC 3 A3 = P A3 B1 + P A3 B2 + P A3 B3 = P A3 f) (Extra Credit) Find the probability that the jorcillator fails in the third period, given that the PC 3 A3 P A3 1 phillinx fails in the third month i.e. P C3 A3 (1). This is P A3 P A3 g) Demonstrate Bayes’ rule by finding P A3 C3 and showing that Bayes’ rule explains the relationship between the conditional probabilities that you have found. (2) [34] P A3 C 3 P A3 . Bayes’ Rule says P A3 C3 PC 3 P A3 PB3 P A3 B3 PA3 C3 PC3 A3 P A3 PC3 1P A3 . Just fill in your numbers. PC3 18 251y0831 4/18/08 3. Assume that you have a variable with a Poisson distribution with a mean of 1.5 .1g . (My value of g is 2, so I have a mean of 1.5 .12 1.5 .2 1.3 . Write down the values of P0 , P 1 etc for all values of x that have probabilities of .0001or larger by subtracting values from the Poisson table. Compute the expected value and standard deviation using the probabilities and values of x and show that these are the same as stated in our write-ups on the Poisson distribution. (4) [37] If g 1, m 1.4 . x P x xPx x 2 P x 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 0.246597 0.345236 0.241665 0.112777 0.039472 0.011052 0.002579 0.000516 0.000090 0.000014 0.000002 1.00000 0.000000 0.345236 0.483330 0.338331 0.157888 0.055260 0.015474 0.003612 0.000720 0.000126 0.000020 1.40000 0.00000 0.34524 0.96666 1.01499 0.63155 0.27630 0.09284 0.02528 0.00576 0.00113 0.00020 3.3600 Px 1 (This shows that we have a valid distribution), xPx 1.40000 and E x x Px 3.36000 . This means that E x 3.36000 1.4000 1.4000 So 2 2 2 If g 9, m 0.6 . x 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 P x 0.548812 0.329287 0.098786 0.019757 0.002964 0.000356 0.000036 0.000003 1.00000 xPx 0.000000 0.329287 0.197572 0.059271 0.011856 0.001780 0.000216 0.000021 0.60000 2 2 2 x 2 P x 0.000000 0.329287 0.395144 0.177813 0.047424 0.008900 0.001296 0.000147 0.9600 Px 1 (This shows that we have a valid distribution), xPx 0.60000 and E x x Px 0.9600 . This means that E x 0.9600 0.6000 0.6000 . So 2 2 2 2 2 2 Rounding error may knock these totals slightly off. xPx 1.3 and E x 2 2.99. So 2 E x 2 2 1.30 If g 2, m 1.3 . If g 3, m 1.2 . xPx 1.2 and Ex 2 2.64. So 2 E x 2 2 1.20 If g 4, m 1.1 . xPx 1.1 and Ex 2 2.31. So 2 E x 2 2 1.10 If g 5, m 1.0 . xPx 1.0 and Ex 2 2.00. So 2 E x 2 2 1.00 If g 6, m 0.9 . xPx 0.9 and Ex 2 1.71. So 2 E x 2 2 0.90 If g 7, m 0.8 . xPx 0.8 and Ex 2 1.44. So 2 E x 2 2 0.80 If g 8, m 0.7 . xPx 0.7 and Ex 2 1.19. So 2 E x 2 2 0.70 19