TEST3_F22_Solutions_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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{ 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] Carol and Stacey are both training for a 5K race and each are hoping to place in the top tier of fin- ishers, where the top tier are all participants who cross the finish line within 30 minutes. The time it takes Carol to run 5K has an exponential distribution with mean 25 minutes. The time it takes Stacey to run 5K has an exponential distribution with mean 28 minutes. Each person completes 6 training runs. Let the random variable X denote the number of times Carol’s 5K training run takes less than 30 minutes, and random variable Y denote the number of times Stacey’s 5K training run takes less than 30 minutes. We assume all training runs are independent (both between Carol and Stacey and for each of the individual’s runs). 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 ≤ 30) where D ∼ Exp ( θ = 25). I.e. p 1 = 1 − e − 30 25 = 0 . 699 to 3 decimal places. For Y the success probability p 2 = P ( D ≤ 30) where D ∼ Exp ( θ = 28). I.e. p 2 = 1 − e − 30 28 = 0 . 657 to 3 decimal places. I.e. X ∼ Bin ( n = 6 , p 1 = 0 . 699) and Y ∼ Bin ( n = 6 , p 2 = 0 . 657). (b) [2 marks] Let f ( x,y ) denote the joint probability function for X and Y . Calculate f (3 , 4). Since all runs are independent, f ( x,y ) = f X ( x ) ∗ f Y ( y ). Thus, f (3 , 4) = P ( X = 3 , Y = 4) = P ( X = 3) ∗ P ( Y = 4). Therefore, f (3 , 4) = ( 6 3 ) (0 . 699) 3 (0 . 301) 3 ∗ ( 6 4 ) (0 . 657) 4 (0 . 343) 2 = 0 . 0612. (c) [2 marks] Let T = X + Y be the random variable denoting the total number of training runs that take less than 30 minutes by Stacey and Carol 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) = 6 2 (0 . 699) 2 (0 . 301) 4 ∗ (0 . 343) 6 + 6 1 (0 . 699) 1 (0 . 301) 5 ∗ 6 1 (0 . 657) 1 (0 . 343) 5 + (0 . 301) 6 ∗ 6 2 (0 . 657) 2 (0 . 343) 4 = 0 . 000359 (d) [2 marks] Let f T | X ( t,x ) denote the conditional probability function for the total training runs under 30 minutes given Carol’s number of training runs under 30 minutes. 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) = 6 2 (0 . 657) 2 (0 . 343) 4 = 0 . 0896

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Question 2 [8 marks] Suppose adults’ thyroid stimulating hormone (TSH) levels follow a Normal distribution with mean 2.2 milli-international units per liter (mIU/L) and variance 0.8 mIU 2 /L 2 . Adults with TSH levels less than 0.4 mIU/L are considered to have an overactive thyroid. (a) [2 marks] Calculate the probability that a randomly selected adult has an overactive thyroid. Let W ∼ N (2 . 2 , 0 . 8) be a random adult’s TSH level. P ( W < 0 . 4) = P Z < 0 . 4 − 2 . 2 √ 0 . 8 = P ( Z < − 2 . 01) = 1 − P ( Z < 2 . 01) = 1 − 0 . 97778 = 0 . 0222 . (b) [2 marks] Calculate the probability that an adult’s TSH level is within 1 mIU/L of the mean. P (1 . 2 < W < 3 . 2) = P 1 . 2 − 2 . 2 √ 0 . 8 < Z < 3 . 2 − 2 . 2 √ 0 . 8 = P ( − 1 . 12 < Z < 1 . 12) = 1 − 2(1 − P ( Z < 1 . 12)) = 1 − 2(1 − 0 . 86864) = 0 . 737 . (c) [2 marks] Calculate the probability an adult’s TSH level is below the mean given the adult does not have an overactive thyroid. P ( W < 2 . 2 | W > 0 . 4) = P ( W < 2 . 2 ∩ W > 0 . 4) /P ( W > 0 . 4) = P (0 . 4 < W < 2 . 2) / 0 . 97778 = P 0 . 4 − 2 . 2 √ 0 . 8 < Z < 2 . 2 − 2 . 2 √ 0 . 8 / 0 . 97778 = P ( − 2 . 01 < Z < 0) / 0 . 97778 = (Φ(2 . 01) − 0 . 5) / 0 . 97778 = (0 . 97778 − 0 . 5) / 0 . 97778 = 0 . 489 . (d) [2 marks] The term “hypothyroidism” is used to describe the condition of individuals who present TSH levels that are too high. Hypothyroidism is thought to affect around 10% of the adult population. Find the minimum TSH level of an adult with hypothyroidism. We need to solve the following equation for x : 0 . 9 = P ( W < x ) = P Z < x − 2 . 2 √ 0 . 8 = Φ( x − 2 . 2 √ 0 . 8 ) . From the Z-table we get 1 . 28 = x − 2 . 2 √ 0 . 8 , which gives a minimum TSH of x = 3 . 345 mIU/L.

Question 3 [8 marks] At a high school, Vice Principal Galecki makes calls to parents/guardians for various reasons, including issues with students (e.g., behaviour, poor grades, and missing class), and administrative reasons. The probabilities of making any such outgoing calls are 0.25 for behaviour, 0.15 for poor grades, 0.30 for missing classes, and 0.30 for administrative reasons. Let X i ( i = 1 , . . . , 4) represent the number of outgoing calls of type behaviour, poor grades, missing class, and administrative respectively, in a set of 18 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 ) = 18! x 1 ! x 2 ! x 3 ! x 4 ! (0 . 25) x 1 (0 . 15) x 2 (0 . 30) x 3 (0 . 30) x 4 where x i = 0 , 1 , . . . , 18, i = 1 , 2 , 3 , 4, and x 1 + x 2 + x 3 + x 4 = 18. (b) In the set of 18 calls, calculate the probability that Vice Principal Galecki makes 5 calls regarding administrative reasons. [2 marks] Since X 4 ∼ Bin (18 , 0 . 30), then P ( X 4 = 5) = ( 18 5 ) (0 . 30) 5 (1 − 0 . 30) 18 − 5 = 0 . 202 (c) In the set of 18 calls, calculate the probability that Vice Principal Galecki makes 12 calls related to student issues. [2 marks] X 1 + X 2 + X 3 ∼ Bin (18 , 0 . 25 + 0 . 15 + 0 . 3 = 0 . 7) P ( X 1 + X 2 + X 3 = 12) = ( 18 12 ) (0 . 7) 12 (0 . 3) 6 = 0 . 187 (d) In the set of 18 calls, Vice Principal Galecki makes 12 calls related to student issues. Calculate the probability that at most one of those calls is regarding poor grades. [2 marks] P ( X 2 ≤ 1 | X 1 + X 2 + X 3 = 12) = P ( X 2 = 0 | X 1 + X 2 + X 3 = 12) + P ( X 2 = 1 | X 1 + X 2 + X 3 = 12) = P ( X 2 = 0 , X 1 + X 2 + X 3 = 12) P ( X 1 + X 2 + X 3 = 12) + P ( X 2 = 1 , X 1 + X 2 + X 3 = 12) P ( X 1 + X 2 + X 3 = 12) = P ( X 2 = 0 , X 1 + X 3 = 12) P ( X 1 + X 2 + X 3 = 12) + P ( X 2 = 1 , X 1 + X 3 = 11) P ( X 1 + X 2 + X 3 = 12) = 18! 0!12!6! (0 . 15) 0 (0 . 25 + 0 . 3) 12 (0 . 3) 6 18! 12!6! (0 . 7) 12 (0 . 3) 6 + 18! 1!11!6! (0 . 15) 1 (0 . 25 + 0 . 3) 11 (0 . 3) 6 18! 12!6! (0 . 7) 12 (0 . 3) 6 = (11 / 14) 12 + 12 0 . 15(0 . 55) 11 0 . 7 12 = 0 . 237

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 (3 , 9) 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 − 3 9 − 3 = u for x , which gives x = 3 + 6 u . Thus, the transformation is h ( u ) = 3 + 6 u . (b) [2 marks] You wish to simulate realizations from a random variable W which has a discrete U (3 , 9) 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 ] − 3+1 9 − 3+1 = [ x ] − 2 7 ≥ u iif [ x ] ≥ 2 + 7 u , the smallest x is ⌈ 2 + 7 u ⌉ . So we need h ( u ) = ⌈ 2 + 7 u ⌉ . Alternatively, the answer could be given in the form of a piecewise function: h ( u ) = 3 if 0 < u ≤ 1 / 7, h ( u ) = 4 if 1 / 7 < u ≤ 2 / 7, h ( u ) = 5 if 2 / 7 < u ≤ 3 / 7, h ( u ) = 6 if 3 / 7 < u ≤ 4 / 7, h ( u ) = 7 if 4 / 7 < u ≤ 5 / 7, h ( u ) = 8 if 5 / 7 < u ≤ 6 / 7, h ( u ) = 9 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 ) = [3 + 7 u ] as a correct answer or h ( u ) = 3 if 0 ≤ u < 1 / 7, h ( u ) = 4 if 1 / 7 ≤ u < 2 / 7, h ( u ) = 5 if 2 / 7 ≤ u < 3 / 7, h ( u ) = 6 if 3 / 7 ≤ u < 4 / 7, h ( u ) = 7 if 4 / 7 ≤ u < 5 / 7, h ( u ) = 8 if 5 / 7 ≤ u < 6 / 7, h ( u ) = 9 if 6 / 7 ≤ u < 1. (c) [2 marks] Suppose tornadoes in a province occur according to a Poisson process with an average of 5 tornadoes per year. You wish to simulate all future tornado occurrences of this Poisson process in the year 2023. Your computer simulates values for U as follows: 0.5843, 0.4033, 0.8766, 0.8714, 0.3457, 0.0225, 0.8202, 0.6407, etc. Use the simulated values for U to obtain days of year (integers between 1 and 365) of all the tornado 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 . 2 year/tornado, h ( U ) = − 0 . 2 log (1 − U ) ∼ Exp (0 . 2) and h (0 . 5843) = 0 . 175558287 , h (0 . 4033) = 0 . 103268161 , h (0 . 8766) = 0 . 418464834 , h (0 . 8714) = 0 . 410209693, etc. Since h ( U ) ∼ Exp (0 . 2) the num- bers we obtained are years between events. So event times in years are 0 . 175558287 , 0 . 175558287 + 0 . 103268161 = 0 . 2788264 , 0 . 2788264 + 0 . 418464834 = 0 . 6972912 , 0 . 6972912 + 0 . 410209693 = 1 . 107501. Only the first 3 events happened within 2023 because the 4th one happens after 1.11 years. Multiplying by 365 and rounding up we get days of events: 65, 102, 255. 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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TEST3_F22_Solutions_B

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