TEST3_F22_Solutions_A

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Jan 9, 2024

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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 30 marks. The number of marks available per question is indicated in [square brackets]. • Answer the questions in the spaces provided. You may use the extra pages at the end 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. • Formula sheets are provided in the last pages of the test. You may detach the pages.

Question 1 [8 marks] At Camp Nowhere , the receptionist makes calls to parents/guardians for various reasons, including issues with campers (e.g., homesickness, injury, and behaviour), and administrative reasons. The probabilities of making any such outgoing calls are 0.50 for homesickness, 0.15 for injury, 0.10 for behaviour, and 0.25 for administrative reasons. Let X i ( i = 1 , . . . , 4) represent the number of outgoing calls of type homesickness, injury, behaviour, and administrative respectively, in a set of 25 calls. Assume outgoing calls are independent. (a) Give the joint probability function for X 1 , X 2 , . . . , X 4 , and it’s range. [2 marks] P ( X 1 = x 1 , . . . , X 4 = x 4 ) = 25! x 1 ! x 2 ! x 3 ! x 4 ! (0 . 5) x 1 (0 . 15) x 2 (0 . 10) x 3 (0 . 25) x 4 where x i = 0 , 1 , . . . , 25, i = 1 , 2 , 3 , 4, and x 1 + x 2 + x 3 + x 4 = 25. (b) In the set of 25 calls, calculate the probability that the receptionist makes 7 calls regarding administrative reasons. [2 marks] Since X 4 ∼ Bin (25 , 0 . 25), then P ( X 4 = 7) = ( 25 7 ) (0 . 25) 7 (1 − 0 . 25) 25 − 7 = 0 . 165 (c) In the set of 25 calls, calculate the probability that the receptionist makes 15 calls related to camper issues. [2 marks] X 1 + X 2 + X 3 ∼ Bin (25 , 0 . 5 + 0 . 15 + 0 . 1 = 0 . 75) P ( X 1 + X 2 + X 3 = 15) = ( 25 15 ) (0 . 75) 15 (0 . 25) 10 = 0 . 0417 (d) In the set of 25 calls, the receptionist makes 15 calls related to camper issues. Calculate the probability that at least two of those calls are regarding camper behaviour. [2 marks] P ( X 3 ≥ 2 | X 1 + X 2 + X 3 = 15) = 1 − P ( X 3 = 0 | X 1 + X 2 + X 3 = 15) − P ( X 3 = 1 | X 1 + X 2 + X 3 = 15) = 1 − P ( X 3 = 0 , X 1 + X 2 + X 3 = 15) P ( X 1 + X 2 + X 3 = 15) − P ( X 3 = 1 , X 1 + X 2 + X 3 = 15) P ( X 1 + X 2 + X 3 = 15) = 1 − P ( X 3 = 0 , X 1 + X 2 = 15) P ( X 1 + X 2 + X 3 = 15) − P ( X 3 = 1 , X 1 + X 2 = 14) P ( X 1 + X 2 + X 3 = 15) = 1 − 25! 0!15!10! (0 . 1) 0 (0 . 5 + 0 . 15) 15 (0 . 25) 10 25! 15!10! (0 . 75) 15 (0 . 25) 10 − 25! 1!14!10! (0 . 1) 1 (0 . 5 + 0 . 15) 14 (0 . 25) 10 25! 15!10! (0 . 75) 15 (0 . 25) 10 = 1 − (13 / 15) 15 − 15 0 . 1(0 . 65) 14 0 . 75 15 = 0 . 613

Question 2 [8 marks] Suppose babies’ birth weights follow a Normal distribution with mean 7 pounds and variance 0.4 pound 2 . Babies less than 5.5 pounds are considered low birth weight. (a) [2 marks] Calculate the probability a baby is born with low birth weight. Let W ∼ N (7 , 0 . 4) be a random baby’s weight. P ( W < 5 . 5) = P Z < 5 . 5 − 7 √ 0 . 4 = P ( Z < − 2 . 37) = 1 − P ( Z < 2 . 37) = 1 − 0 . 99111 = 0 . 00889 . (b) [2 marks] Calculate the probability a newborn’s weight is within 1 pound of the mean. P (6 < W < 8) = P 6 − 7 √ 0 . 4 < Z < 8 − 7 √ 0 . 4 = P ( − 1 . 58 < Z < 1 . 58) = 1 − 2(1 − P ( Z < 1 . 58)) = 1 − 2(1 − 0 . 94295) = 0 . 886 . (c) [2 marks] Calculate the probability a newborn’s weight is below the mean given the newborn does not have a low birth weight. P ( W < 7 | W > 5 . 5) = P ( W < 7 ∩ W > 5 . 5) /P ( W > 5 . 5) = P (5 . 5 < W < 7) / 0 . 99111 = P 5 . 5 − 7 √ 0 . 4 < Z < 7 − 7 √ 0 . 4 / 0 . 99111 = P ( − 2 . 37 < Z < 0) / 0 . 99111 = (Φ(2 . 37) − 0 . 5) / 0 . 99111 = (0 . 99111 − 0 . 5) / 0 . 99111 = 0 . 496 . (d) [2 marks] The term “fetal macrosomia” is used to describe newborns who are much larger than average. Around 9% of babies have fetal macrosomia. Find the minimum weight of a newborn with fetal macrosomia. We need to solve the following equation for x : 0 . 91 = P ( W < x ) = P Z < x − 7 √ 0 . 4 = Φ( x − 7 √ 0 . 4 ) . From the Z-table we get 1 . 34 = x − 7 √ 0 . 4 , which gives a minimum weight of x = 7 . 847 pounds.

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Question 3 [8 marks] Eric and Jared are playing a carnival game where you roll a marble on the floor and measure the distance the marble travels. Marbles which travel a distance of at least 4 meters win a prize. The distance Eric’s marble travels has an exponential distribution with mean 2.8 meters. The distance Jared’s marble travels has an exponential distribution with mean 3.0 meters. Each person is given 5 chances to roll their marble and they win a prize for each of the 5 rolls that pass the 4 meter mark. Let the random variable X denote the number of times Eric’s marble travels at least 4 meters, and random variable Y denote the number of times Jared’s marble travels at least 4 meters. We assume all marbles rolls are independent (both between Eric and Jared and for each individuals’ rolls). Reminder: F ( x ) = 1 − e − x θ for x ≥ 0 when X ∼ Exp ( θ ) (a) [2 marks] What are the marginal distributions for X and Y ? Identify the name and parameter values (to 3 decimal places) for each of X and Y ’s distributions. Use the rounded parameter values found here for parts b)c)d). Both X and Y have Binomial distributions. For X the success probability p 1 = P ( D > 4) where D ∼ Exp ( θ = 2 . 8). I.e. p 1 = e − 4 2 . 8 = 0 . 240 to 3 decimal places. For Y the success probability p 2 = P ( D > 4) where D ∼ Exp ( θ = 3 . 0). I.e. p 2 = e − 4 3 = 0 . 264 to 3 decimal places. I.e. X ∼ Bin ( n = 5 , p 1 = 0 . 240) and Y ∼ Bin ( n = 5 , p 2 = 0 . 264). (b) [2 marks] Let f ( x,y ) denote the joint probability function for X and Y . Calculate f (2 , 1). Since all rolls are independent, f ( x,y ) = f X ( x ) ∗ f Y ( y ). Thus, f (2 , 1) = P ( X = 2 , Y = 1) = P ( X = 2) ∗ P ( Y = 1). Therefore, f (2 , 1) = ( 5 2 ) (0 . 240) 2 (1 − 0 . 240) 3 ∗ ( 5 1 ) (0 . 264) 1 (1 − 0 . 264) 4 = 0 . 0979. (c) [2 marks] Let T = X + Y be the random variable denoting the total number of prizes won by Eric and Jared combined. Find P ( T = 2).. Reminder that we cannot use the result T = X + Y ∼ Bin ( n + m, p ) when X and Y do not have have a common success probability p . Thus, for P ( T = 2), we consider all possible cases where X + Y = 2 That is, P ( T = 2) = P ( X = 2 , Y = 0) + P ( X = 1 , Y = 1) + P ( X = 0 , Y = 2) = 5 2 (0 . 240) 2 (0 . 76) 3 ∗ (0 . 736) 5 + 5 1 (0 . 240) 1 (0 . 76) 4 ∗ 5 1 (0 . 264) 1 (0 . 736) 4 + (0 . 76) 5 ∗ 5 2 (0 . 264) 2 (0 . 736) 3 = 0 . 280 (d) [2 marks] Let f T | X ( t,x ) denote the conditional probability function for the total prizes won given the number of prizes Eric wins. Calculate f T | X (2 , 0).

f T | X (2 , 0) = P ( T = 2 , X = 0) P ( X = 0) = P ( X = 0 , Y = 2) P ( X = 0) = P ( X = 0) P ( Y = 2) P ( X = 0) = P ( Y = 2) = 5 2 (0 . 264) 2 (0 . 736) 3 = 0 . 278

Question 4 [6 marks] A computer is able to simulate realizations from a continuous random variable U ∼ U (0 , 1). (a) [2 marks] You wish to simulate realizations from a random variable Y which has a continuous U (2 , 8) distribution. Find a transformation h such that Y = h ( U ). Hint: We showed in class that the CDF of X ∼ U ( a,b ) is F X ( x ) = x − a b − a when a ≤ x ≤ b . We find F − 1 by solving = x − 2 8 − 2 = u for x , which gives x = 2 + 6 u . Thus, the transformation is h ( u ) = 2 + 6 u . (b) [2 marks] You wish to simulate realizations from a random variable W which has a discrete U (2 , 8) distribution. Find a transformation h such that W = h ( U ). Hint: We showed in class that the CDF of X ∼ discrete U ( a,b ) is F ( x ) = [ x ] − a +1 b − a +1 when a ≤ x < b . We need to find the generalized inverse F − 1 ( x ) = inf( x : F ( x ) ≥ u ). Noting that F ( x ) ≥ u iif [ x ] − 2+1 8 − 2+1 = [ x ] − 1 7 ≥ u iif [ x ] ≥ 1 + 7 u , the smallest x is ⌈ 1 + 7 u ⌉ . So we need h ( u ) = ⌈ 1 + 7 u ⌉ . Alternatively, the answer could be given in the form of a piecewise function: h ( u ) = 2 if 0 < u ≤ 1 / 7, h ( u ) = 3 if 1 / 7 < u ≤ 2 / 7, h ( u ) = 4 if 2 / 7 < u ≤ 3 / 7, h ( u ) = 5 if 3 / 7 < u ≤ 4 / 7, h ( u ) = 6 if 4 / 7 < u ≤ 5 / 7, h ( u ) = 7 if 5 / 7 < u ≤ 6 / 7, h ( u ) = 8 if 6 / 7 < u ≤ 1. Alternate solution: since the exact values 1/7, 2/7, ..., 6/7 have probability 0 to be simulated, we can also accept h ( u ) = [2 + 7 u ] as a correct answer or h ( u ) = 2 if 0 ≤ u < 1 / 7, h ( u ) = 3 if 1 / 7 ≤ u < 2 / 7, h ( u ) = 4 if 2 / 7 ≤ u < 3 / 7, h ( u ) = 5 if 3 / 7 ≤ u < 4 / 7, h ( u ) = 6 if 4 / 7 ≤ u < 5 / 7, h ( u ) = 7 if 5 / 7 ≤ u < 6 / 7, h ( u ) = 8 if 6 / 7 ≤ u < 1. (c) [2 marks] Suppose methane leaks at an oil and gas facility occur according to a Poisson process with an average of 4 leaks per year. You wish to simulate all future leak occurrences of this Poisson process in the year 2023. Your computer simulates values for U as follows: 0.8766, 0.0525, 0.3457, 0.8714, 0.8202, 0.5843, 0.6407, 0.4033, etc. Use the simulated values for U to obtain days of year (integers between 1 and 365) of all the leak occurrences in 2023. Start simulating from the beginning of 2023, using the simulated values for U in the order presented above. Hint: We showed in class that − θlog (1 − U ) ∼ Exp ( θ ), where log stands for the natural logarithm. We have θ = 0 . 25 year/leak, h ( U ) = − 0 . 25 log (1 − U ) ∼ Exp (0 . 25) and h (0 . 8766) = 0 . 52308104 , h (0 . 0525) = 0 . 01348209 , h (0 . 3457) = 0 . 10604733 , h (0 . 8714) = 0 . 51276212, etc. Since h ( U ) ∼ Exp (0 . 25) the num- bers we obtained are years between events. So event times in years are 0 . 52308104 , 0 . 52308104 + 0 . 01348209 = 0 . 5365631 , 0 . 5365631 + 0 . 10604733 = 0 . 6426104 , 0 . 6426104 + 0 . 51276212 = 1 . 155373. Only the first 3 events happened within 2023 because the 4th one happens after 1.15 years. Multiply- ing by 365 and rounding up we get days of events: 191, 196 and 235. 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.

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f x E X Var X M t a b b a b a a x b a b b a e bt e at b a t t t e x x t t N G e x x e t t t

TEST3_F22_Solutions_A

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