Test2_SOLUTIONS_F22_STAT230_A

{ 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 [8 marks] MacDonald’s is a small town restaurant that specializes in making burgers. In general, their stockroom contains 50 condiment packets of which 25 are ketchup, 15 are mustard, and 10 are relish. The manager randomly draws 20 packets from the stockroom without replacement to keep at the cash register. (a) [2 marks] Let X be the number of packets drawn that are not relish. Give the probability function for X . (Be sure to include the range for X ). f ( x ) = P ( X = x ) = ( 40 x )( 10 20 − x ) ( 50 20 ) , x = 10 , . . . , 20 (b) [1 mark] Calculate the probability that there are exactly 17 packets at the cash register that are not relish. P ( X = 17) = ( 40 17 )( 10 20 − 17 ) ( 50 20 ) = ( 40 17 )( 10 3 ) ( 50 20 ) = 0 . 226 (c) [2 marks] After running out of all their condiment packets, the manager quickly orders and receives a shipment of 1,000 packets of which 20% are ketchup, 20% are mustard, and 60% are relish. Again, the manager randomly draws 20 packets from the shipment without replacement to keep at the cash register. Calculate the exact probability that at least three of the packets drawn are not relish. P ( X ≥ 3) = 1 − P ( X < 3) = 1 − ( 400 0 )( 600 20 ) + ( 400 1 )( 600 19 ) + ( 400 2 )( 600 18 ) ( 1000 20 ) = 0 . 997 (d) [3 marks] Approximate the probability in part (c) using a suitable approximation. Justify the approximation. Since n = 20 draws is small compared to N = 1000 packets and r = 400 packets, this probability can be approximated using the Binomial approximation. P ( X ≥ 3) ≈ 1 − ∑ 2 x =0 ( 20 x ) (0 . 4) x (0 . 6) 20 − x = 1 − (0 . 6) 20 − 20(0 . 4)(0 . 6) 19 − 190(0 . 4) 2 (0 . 6) 18 = 0 . 996

(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 the delivery of Amazon packages. Carefully justify your opinions. Independence : This seems reasonable since one would expect the packages delivered to one address won’t influence the number of packages delivered to other addresses. Individuality : This is likely not satisfied as it possible for a single address/household to receive two separate Amazon packages at once. Homogeneity : This is also likely not satisfied since different geographical areas would have different delivery rates. You would expect high density residential areas like condo buildings or subdivisions to have a higher rate of package delivery per hour than when the driver gets to the more rural areas of their route. 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.

Question 3 [8 marks] At a Fall fair, contestants are offered the opportunity to play a game that costs 6 tokens to play. Contestants get to toss 3 fair coins and win a number of tokens equal to X 2 +1 where X is the number of heads across the 3 tosses. (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 ( TTT ) = 1 / 8 while P ( X = 1) = P ( HTT ) + P ( THT ) + P ( TTH ) = 3 / 8. (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 − 6 = X 2 − 5 be the net winnings. We have E ( Y ) = (0 2 − 5)(1 / 8) + (1 2 − 5)(3 / 8) + (2 2 − 5)(3 / 8) + (3 2 − 5)(1 / 8) = − 2 . (d) [2 marks] At the next Fall fair, the owner would like to at least double his expected profit. Assuming attendance at the next Fall fair decreases by 25% while every other parameter remains constant, how should the owner adjust the cost of the game (in tokens) to meet his expected profit? From c), the owner’s expected profit is 2 tokens per customer in the current year. With n customers in the current year, the expected profit is 2 n . The following year, the expected profit is thus 2 n (0 . 75) = 1 . 5 n which he wants to increase to 2 n (2) = 4 n . He thus needs to collect an extra 4 n − 1 . 5 n = 2 . 5 n to meet his target. Dividing this into 0 . 75 n customers, the cost of the game must increase by 2 . 5 / 0 . 75 = 3 . 333 tokens to meet his target. And thus he will need to charge an extra 4 tokens per customer (or a total of 10).

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.