mat2378-assignment2-with_sol

mat2378-assignment2-with_sol - Assignment 3 Total number of...

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Unformatted text preview: Assignment 3 Total number of points: 33 Q1. The three events are shown on the Venn diagram: ✬✩ ✬✩ ✫✪ ✫✪ ✬✩ ✫✪ (b) (d) C A B Due date: 7 October 2009 Reproduce the figure and shade the region corresponding to the following events: (a) (c) (e) Ac (A and B ) or C (A and B )c and C (A and B ) or (A and B c ) (B or C )c Solution to Q1: Marking scheme for Q1: Correct answer for each part - 1 point. Total - 5 points. Q2. (Q3.6) In a certain college, 55% of the students are women. We take a sample of two students. Find the probability that (a) both are women; (b) at least on of the two students is a woman. Solution to Q2: (a) (b) 0.55 × 0.55 = 0.3025 0.55 × 0.55 + 0.45 × 0.55 + 0.45 × 0.55 = 0.7975 Marking scheme for Q2: Correct answer for each part - 1 point. Total - 2 points. Q3. If P (A) = 0.3, P (B ) = 0.2, and P (A ∩ B ) = 0.1, determine the following probabilities: (a) P (Ac ) (b) (e) P (A or B ) (c) (f) (d) P (A and B c ) P ((A or B )c ) P (Ac and B ) P (Ac or B ) Solution to Q3: (a) (b) (c) (d) P (Ac ) = 1 − P (A) = 0.7 P (A or B ) = P (A) + P (B ) − P (A and B ) = 0.4 P (A and B c ) = P (A) − P (A and B ) = 0.2 P (Ac and B ) = P (B ) − P (A and B ) = 0.1 (draw a Venn diagram to see this) (e) (f) P ((A or B )c ) = 1 − P (A or B ) = 1 − 0.4 = 0.6 P (Ac or B ) = P (Ac ) + P (B ) − P (Ac and B ) = 0.7 + 0.2 − 0.1 = 0.8 Marking scheme for Q3: Correct answer for each part - 1 point. Total - 6 points. Q4. In a study of the relationship between health risk and income, a large group of people were asked a series of questions. Some of the responses are shown in the following table: Low Smoke 634 Don’t smoke 1846 (a) (b) (c) Income Medium High 332 247 1622 1868 What is the probability that someone in this study smokes? What is the conditional probability that someone in this study smokes, given that the person has high income? Is being a smoker independent of having a high income? Why or why not? Solution to Q4: The full table is Income Low Medium High Total Smoke 634 332 247 1213 Don’t smoke 1846 1622 1868 5336 Total 2480 1954 2115 6549 Let E denotes the event that a person smokes, H - event that a person has a high income. (a) P (E ) = 1213/6549 = 0.185 (b) (c) P (E |H ) = 247/2115 = 0.117 No, since P (E ) ￿= P (E |H ). Marking scheme for Q4: Correct answer for each part - 1 point. Total - 3 points. Q5. In 1988, Physicians’ Health Study Research Group released the results of a five-year experiment conducted using 22,071 male physicians between the ages of 40 and 84. The purpose was to determine whether taking aspirin reduces the risk of a heart attack. The physicians had been randomly assigned to one of the treatment groups. On group took an aspirin, the other took placebo. The following information has been obtained. Aspirin Placebo One (a) (b) (c) (d) # with heart attack # without heart attack 104 10, 933 189 10, 845 person is selected by chance. Let A be an event that a randomly selected person has had a heart attack. Find P (A). Let B be an event that a randomly selected person took aspirin. Find P (B ). Find the probability that it has a heart attack given that it was in the Aspirin group. Are A and B independent? Solution to Q5: total # with heart attack 293 = = 0.013 22, 071 22, 071 11, 037 (b) P (B ) = = 0.5 22, 071 P (A ∩ B ) 104/22, 071 (c) P (A|B ) = = = 0.01 P (B ) 0.5 (d) We have P (A) > P (A|B ), thus A and B are not independent. This study suggests that taking aspirin reduces a heart attack risk. (a) P (A) = Marking scheme for Q5: Correct answer for each part - 1 point. Total - 4 points. Q6. The sample space of a random experiment is {a, b, c, d, e, f } and each outcome is equally likely. A random variable is defined as follows outcome a b c d ef x 0 0 1.5 1.5 2 3 Determine the probability distribution function of Y . Determine the following probabilities: (a) (d) P (Y = 1.5) P (0 ≤ Y < 2) (b) (e) P (0.5 < Y < 2.7) P (Y = 0 or Y = 2) (c) P (Y > 3) Solution to Q6: Probability mass function is P (Y = 0) = P ({a, b}) = (a) (c) (e) P (Y = 1.5) = P (Y > 3) = 0 P (Y = 0 or Y = 2) = 3 6 2 6 11 2 2 1 1 + = ; P (Y = 1.5) = , P (Y = 2) = , P (Y = 3) = . 66 6 6 6 6 (b) (d) P (0.5 < Y < 2.7) = P (Y = 1.5) + P (Y = 2) = P (0 ≤ Y < 2) = P (Y = 0) + P (Y = 1.5) = 4 6 3 6 Marking scheme for Q6: Completely correct p.d.f. - 1 point, correct answer for each part - 1 point. Total - 6 points. Q7. Determine the mean and the variance in Question Q6. Solution to Q7: E(Y ) = 1.33, Var(Y ) ≈ 1.15 Marking scheme for Q7: 1 point for the mean and the variance. Total - 2 points. Q8. Samples of rejuvenated mitochondria are mutated in 1% cases. Suppose 15 samples are studied and they can be considered to be independent for mutation. Determine the following probabilities: (a) No samples are mutated (b) (c) At most one sample is mutated More than half the samples are mutated Notes to Q8: Check your answer using MINITAB. See the webpage for MINITAB help. Solution to Q8: Let Y be the number of no-mutated samples; then Y has binomial distribution with n = 15 and p = 0.99 (success=no mutation; note that if choose success=mutation, then p = 0.01). To compute (a) 0 mutated samples = 15 no-mutated samples, thus to compute P (Y = 15) ≈ 0.86 (b) (c) at most one sample mutated = at least 14 are not mutated; P (Y ≥ 14) = P (Y = 14) + P (Y = 15) ≈ 0.9904 More than half the samples are mutated = less than or half samples are no-mutated; P (Y ≤ 7.5) = P (Y ≤ 7) = 0. Marking scheme for Q8: Correct answer for each part - 1 point. Total - 3 points. Q9. Comment on a shape of probability distribution function of B (30, 0.1). USE MINITAB to answer this question. Include a plot. Marking scheme for Q9: Correct answer - 1 point. Plot included - 2 points. Total - 3 points. ...
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This note was uploaded on 01/13/2010 for the course MATH MATH2378 taught by Professor Kulik during the Spring '09 term at University of Ottawa.

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