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Unformatted text preview: Assignment 3
Total number of points: 32
Q1. A medical research team wished to evaluate a proposed screening test for Alzheimer’s disease. The test was given to a random sample of 450 patients with Alzheimer’s disease, in 436 cases the test result was positive. Also, the test was given to a random sample of 500 patients without the disease, only in 5 cases the result was positive. It is known that in the US 11.3% of the population aged 65 and over have Alzheimer’s disease. Find the probability that a person has the disease given that the test was positive. Solution to Q1: A  test positive, D  a person has disease. Given: P (AD) = P (DA) = 436 5 , P (ADc ) = , P (D) = 0.113. To ﬁnd: 450 500 Due date: 21 October 2009 P (DA) P (A and D) P (A) 436 We have P (A and D) = P (AD)P (D) = 0.113 = 0.1094844 450 Compute ﬁrst P (A) = P (AD)P (D) + P (ADc )P (Dc ) = The ﬁnal answer is P (DA) = 0.925 Marking scheme for Q1: Total  2 points 436 5 0.113 + (1 − 0.113) = 0.1183544 450 500 Q2. (3.43) A certain drug causes kidney damage in 1% of patients. Suppose the drug is to be tested on 50 patients. Find the probability that (a) none of the patients will experience kidney damage; (b) one or more of the patients will experience kidney damage. Check your answers using MINITAB. Solution to Q2: If Y  number of patients that will experience kidney damage; Y ∼ B (50, 0.01). (a) 0.9950 = 0.6050 (b) 0.3950 Marking scheme for Q2: Correct answer for each part  1 point. Total  2 points. Q3. Assume X ∼ N (5, 42 ). Determine the following: (a) P (X < 11) (b) (d) P (−2 < X < 9) (e) Check your answers using MINITAB. Solution to Q3: P (X > 0) P (2 < X < 8) (c) P (3 < X < 7) (a) Let Z ∼ N (0, 1). P (X < 11) = P (b) (c) (d) (e) X −5 0−5 P (X > 0) = 1 − P (X < 0) = 1 − P < = 1 − P (Z < −1.25) = 1 − 0.10565 = 0.89435 4 4 X −5 7−5 X −5 3−5 P (3 < X < 7) = P (X < 7) − P (X < 3) = P < −P < = P (Z < 4 4 4 4 5) − P (Z < −0.5) = 0.691642 − 0.308538 = 0.3829249 X −5 9−5 X −5 −2 − 5 P (−2 < X < 9) = P (X < 9) − P (X < −2) = P < −P < = P (Z < 4 4 4 4 1) − P (Z < −1.75) = 0.8413 − 0.0401 = 0.8012 X −5 8−5 X −5 2−5 P (2 < X < 8) = P (X < 8) − P (X < 2) = P < −P < = P (Z < 4 4 4 4 0.75) − P (Z < −0.75) = 0.7734 − 0.2266 = 0.5468 X −5 11 − 5 < 4 4 = P (Z < 1.5) = 0.933193 Marking scheme for Q3: Correct answer for each part  1 point. Total  5 points. Q4. (4.9  modiﬁed) The serum cholesterol levels of 17yearolds follow a normal distribution with mean 176 mg/dLi and standard deviation of 30 mg/dLi. What percentage of 17yearolds have serum cholesterol values: (a) (c) (e) 186 or more? 216 or less? between 186 and 216? (b) (d) (f) 156 or less? 121 or more? between 121 and 156? (g) between 156 and 186? Check your answers using MINITAB. Solution to Q4: Let Y ∼ N (176, 302 ). To compute (a) P (Y ≥ 186) = 0.3707 or 37.07% (b) (c) (d) (e) (f) (g) P (Y ≤ 156) = 0.2514 P (Y ≤ 216) = 0.9082 P (Y ≥ 121) = 0.9664 P (186 ≤ Y ≤ 216) = 0.2789 P (121 ≤ Y ≤ 156) = 0.2178 P (156 ≤ Y ≤ 186) = 0.3779. Marking scheme for Q4: Correct answer for each part  1 point. Total  7 points. Q5. If Y ∼ N (5, 4), ﬁnd the value of c such that (a) P (Y ≥ c) = 0.1446 Solution to Q5: (b) P (Y  ≥ c) = 0.3174 (a) (b) P (Y ≥ c) = 0.1446, so that P (Y ≤ c) = 0.8554. Thus Y −5 c−5 c−5 0.8554 = P (Y ≤ c) = P ≤ = P (Z ≤ ). 2 2 2 c−5 Thus, = 1.06 and c = 7.12. 2 Do not do this Marking scheme for Q5: Correct answer for each part (a)  2 points. Total  4 points. Q6. (5.15, continuation of Q4). (a) What percentage of the 17yearolds have serum cholesterol values between 166 and 186? (b) (c) ¯ If Y represents the mean cholesterol value of a random sample of nine 17yearolds from the population, ¯ what is P (166 ≤ Y ≤ 186)? ¯ If Y represents the mean cholesterol value of a random sample of twenty ﬁve 17yearolds from the ¯ population, what is P (166 ≤ Y ≤ 186)? Solution to Q6: Let Y ∼ N (176, 302 ). (a) Here, we have to compute 166 − 176 Y − 176 186 − 176 P (166 ≤ Y ≤ 186) = P < < = P (−0.33 < Z < 0.33) = 0.2586. 30 30 30 302 2 ¯ (b) Here, we have Y ∼ N (µY , σY ) = N 176, = N (176, 100). We compute ¯ ¯ 9 ¯ ¯ ≤ 186) = P 166 − 176 < Y − 176 < 186 − 176 = P (−1 < Z < 1) = 0.6826. P (166 ≤ Y 10 10 10 302 2 ¯ (c) Here, we have Y ∼ N (µY , σY ) = N 176, = N (176, 36). We compute ¯ ¯ 25 ¯ ¯ ≤ 186) = P 166 − 176 < Y − 176 < 186 − 176 = P (−1.67 < Z < 1.67) = 0.9525−0.0475 = 0.905. P (166 ≤ Y 6 6 6 Marking scheme for Q6: Correct answer for part (a)  1 point; for parts (b) and (c)  2 points (1 point for the correct standardization, 1 point for the correct answer). Total  5 points. Q7. The amount of time that a customer spends waiting at an airport checkin counter is a random variable with mean µ = 8.2 minutes and standard deviation σ = 1.5 minutes. Suppose that a random sample of n = 49 customers is taken. Compute the approximate probability that the average waiting time for these customers is: (a) Less than 8.5 minutes. (b) Between 7.9 and 8.5 minutes. Solution to Q7: Let Y be the mean waiting time for 49 clients. Then according to Central Limit Theorem, Y ∼ N (8.2, (1.5)2 /49). (a) 8.5 − 8.2 √ P (Y < 8.5) ≈ P Z < = P (Z < 1.4) = 0.9192. 1.5/ 49 (b) P (7.9 < Y < 8.5) ≈ P 8.5 − 8.2 7.9 − 8.2 √ √ −P Z < 1.5/ 49 1.5/ 49 = P (Z < 1.4) − P (Z < −1.93) = 0.8385. Z< Marking scheme for Q7: For each part: 1 point for the correct standardization, 1 point for the correct answer. Total  4 points. Q8. Download data lentil.mtp. Assess normality. Solution to Q8: Data do not follow normal distribution. Marking scheme for Q8: Correct plots  1 point each. Correct conclusion  1 point. Total  3 points. ...
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This note was uploaded on 01/13/2010 for the course MATH MAT2378 taught by Professor Kulik during the Spring '09 term at University of Ottawa.
 Spring '09
 Kulik

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