Test2_SOLUTIONS_F22_STAT230_B

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University of Waterloo *

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230

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Statistics

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Jan 9, 2024

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9

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{ Special instructions Your final numerical answers should be given to 3 decimal places (e.g. 0.329). However, between steps/parts you should carry more decimal places to avoid rounding errors. If working is not specifically requested, a correct answer will receive full marks. However, an incorrect answer could still receive credit if working is provided. An incorrect answer with no working will receive zero marks. The exam consists of 4 questions for a total of 32 marks. The number of marks available per question is indicated in [square brackets]. Answer the questions in the spaces provided. You may use the last page of the test for any additional work you would like to be graded. If you use this space, make it as clear as possible which question(s) your work relates to.
Question 1 [10 marks] A police officer parks in a discrete location where they can track the speed of cars that drive past with a radar gun. It is assumed that speeding drivers pass the police officer’s location according to a Poisson process at a rate of 3 cars/hour. (a) [2 marks] Calculate the probability that fewer than 4 speeding cars pass the police officer in a one hour period. We have λ = 3 speeding cars/hour for this Poisson process. Letting random variable X = the number of speeding cars in a one hour period, X Poi ( µ = 3). P ( X < 4) = P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) = e 3 + e 3 (3) + e 3 (3) 2 2! + e 3 (3) 3 3! = 0 . 647 (b) [2 marks] Calculate the probability 25 speeding cars pass the police officer over a 7.5-hour work day. Letting random variable Y = the number of speeding cars in a 7.5-hour workday, Y Poi ( µ ) where µ = λ (7 . 5) = 22 . 5 P ( Y = 25) = e 22 . 5 (22 . 5) 25 25! = 0 . 0695 (c) [3 marks] The police officer takes a break once they’ve seen at least 4 speeding cars in each of two separate one-hour periods. Calculate the probability the officer takes a break at the end of their 5 th hour of work. Define success as a one-hour period with at least 4 speeding cars. Then, we have p = probability of success = 1 - 0.647 = 0.353 from part (a). Let random variable Z = the number of hours of work at which point the officer takes a break. We see that Z 2 NB ( k = 2 ,p = 0 . 353). We want P ( Z = 5) = P ( Z 2 = 3). Thus, P ( Z = 5) = 4 3 (0 . 353) 2 (1 0 . 353) 3 = 0 . 135
(d) [3 marks] What are the 3 conditions that must be satisfied for the assumption of a Poisson process to be appropriate? For each condition you list, discuss whether or not you think this assumption is appropriate for speeding cars on a stretch of road. Carefully justify your opinions. Independence : This seems reasonable since one would expect the number of speeding cars in a cer- tain period of time does not influence the number of speeding cars earlier or later in the day. Individuality : This is likely not satisfied as it possible for multiple cars to be speeding side by side at the same time. Homogeneity : This is also likely not satisfied since it is possible drivers are more likely to speed at certain times of day (for example, when the roads are less busy). Marking Note: There are no single right answers for whether or not each assumption is valid or not. Students can still get full marks with different reasoning for their justifications, or even if they differ from the solution above regarding whether a specific assumption is valid or not. As long as their explanation seems plausible and well-justified, it can be marked correct.
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Question 2 [8 marks] At Mr Goose’s Theme Park, contestants are offered the opportunity to play a game that costs 5 tickets to play. Contestants get to roll 3 dice and win a number of tickets equal to X 2 + 1 where X is the number of even die rolls out of the 3 rolls. (a) [2 marks] Does X have a discrete uniform distribution? Carefully justify why or why not. It does not have a uniform distribution because the outcomes are not equally likely. For example, P ( X = 0) = P ( FFF ) = 1 / 8 while P ( X = 1) = P ( SFF ) + P ( FSF ) + P ( FFS ) = 3 / 8 (noting that S denotes an even die roll and F an uneven one). (b) [2 marks] Tabulate the probability function for X and find the expected value of X . We have the following probability function for X : x 0 1 2 3 f ( x ) 1/8 3/8 3/8 1/8 So E ( X ) = 0(1 / 8) + 1(3 / 8) + 2(3 / 8) + 3(1 / 8) = 1 . 5 (c) [2 marks] Find the expected value of a contestant’s net winnings. Let Y = g ( X ) = X 2 + 1 5 = X 2 4 be the net winnings. We have E ( Y ) = (0 2 4)(1 / 8) + (1 2 4)(3 / 8) + (2 2 4)(3 / 8) + (3 2 4)(1 / 8) = 1 . (d) [2 marks] Next year, the owner (Mr Goose) would like to at least triple his expected profit from the game. Assuming attendance in the next year decreases by 20% while every other parameter remains constant, how should the owner adjust the cost of the game (in tickets) to meet his expected profit? From c), the owner’s expected profit is 1 ticket per customer in the current year. With n customers in the current year, the expected profit is n . The following year, the expected profit is thus 0 . 8 n which he wants to increase to 3 n . He thus needs to collect an extra 3 n 0 . 8 n = 2 . 2 n to meet his target. Dividing this into 0 . 8 n customers, the cost of the game must increase by 2 . 2 / 0 . 8 = 2 . 75 tickets to meet his target. And thus he will need to charge an extra 3 tickets per customer (or a total of 8).
Question 3 [8 marks] Taco Tell is a small town restaurant that offers a variety of Mexican-inspired foods, including tacos, burritos, nachos, and so on. The back room contains 800 bags of nacho chips, 10% of which are stale. The manager randomly draws 30 bags of nacho chips from the back room without replacement and determines if they are stale. (a) [2 marks] Let X be the number of stale bags of nacho chips found. Give the probability function for X . (Be sure to include the range for X ) f ( x ) = P ( X = x ) = ( 80 x )( 720 30 x ) ( 800 30 ) , x = 0 , . . . , 30 (b) [1 mark] Calculate the probability that there are exactly 5 stale bags of nacho chips found. P ( X = 5) = ( 80 5 )( 720 30 5 ) ( 800 30 ) = ( 80 5 )( 720 25 ) ( 800 30 ) = 0 . 102 (c) [2 marks] Calculate the probability that at least four stale bags of nacho chips are found. P ( X 4) = 1 P ( X < 4) = 1 ( 80 0 )( 720 30 ) + ( 80 1 )( 720 29 ) + ( 80 2 )( 720 28 ) + ( 80 3 )( 720 27 ) ( 800 30 ) = 0 . 352 (d) [3 marks] Approximate the probability in part (c) using a suitable approximation. Justify the approximation. Since n = 30 draws is small compared to N = 800 bags and r = 80 stale bags, this probability can be approximated using the Binomial approximation. P ( X 4) 1 3 x =0 ( 30 x ) (0 . 1) x (0 . 9) 30 x = 1 (0 . 9) 30 30(0 . 1)(0 . 9) 29 435(0 . 1) 2 (0 . 9) 28 4060(0 . 1) 3 (0 . 9) 27 = 0 . 353
Question 4 [6 marks] Over the course of 6 months (183 days), youngster Jeff challenges his dad 10 times each day at thumb war. Jeff’s probability of winning a game is 2% for each game. You may assume games are independent. (a) [2 marks] Calculate the probability that Jeff wins at least two games on his first day against his dad. 1-P(no win on first day)-P(1 win on first day) = 1- ( 10 0 ) 0 . 02 0 0 . 98 10 ( 10 1 ) 0 . 02 1 0 . 98 9 = 0 . 0162 . (b) [2 marks] Approximate the probability that Jeff wins 40 games over 6 months. The total number of games is 183(10)=1830. So we want: ( 1830 40 ) 0 . 02 40 0 . 98 1790 e 1830 0 . 02 (1830 0 . 02) 40 40! = 0 . 0540 . (c) [2 marks] Calculate the probability that Jeff’s 4th win against his dad occurs on his last game of the 3rd day. Let X NB (4 , 0 . 02) be the number of losses before the 4th win. We want P ( X = 10 3 4) = ( 29 3 ) (0 . 98) 26 (0 . 02) 4 = 0 . 000346 . (d) [ BONUS - 2 marks available] If Jeff wins at least one game in the first week against his dad, calculate the probability that it takes at least 4 days before Jeff wins his first game. P ( 4 days before win | ≥ 1 win in 70 games) = P ( 40 games before win | ≥ 1 win in 70 games) = P (between 41 and 69 games before win) P ( 1 win in 70 games) = P (between 41 and 69 games before win) 1 P (no win in 70 games) = F (69) F (40) 1 0 . 98 70 = [1 (1 0 . 02) 70 ] [1 (1 0 . 02) 41 ] 1 0 . 98 70 = 0 . 256 , using the CDF of the geometric distribution in the numerator. END OF EXAMINATION
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Use this page for any additional work you would like to be graded. If you use this space, make it as clear as possible which question(s) your work relates to. If there is any ambiguity only your work on the previous question pages will be graded.
Use this page for any additional work you would like to be graded. If you use this space, make it as clear as possible which question(s) your work relates to. If there is any ambiguity only your work on the previous question pages will be graded.
Use this page for any additional work you would like to be graded. If you use this space, make it as clear as possible which question(s) your work relates to. If there is any ambiguity only your work on the previous question pages will be graded.
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