Final Exam for Statistics

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MAT 2377C Final exam

December 10, 2010 Time: 180 minutes Student Number: Family Name: First Name:

Professor: Rafal Kulik

This is an open book examination. Only non-programmable and non-graphic calculators are permitted. Record your answer to each question in the table below. Number of pages: 7 (including this one). Number of questions: 24. NOTE: At the end of the examination, hand in only this page. You may keep the questionnaire.

Question 1 2 3 4 5 6 7 8 9 10 11 12

Answer

Question 13 14 15 16 17 18 19 20 21 22 23 24

Answer

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Q1. Suppose that for a very large shipment of integrated-circuit chips, the probability of failure for any one chip is 0.09. Find the probability that at most 2 chips fail in a random sample of size 20. (The numbers are rounded down to the second decimal place). (a) 0.13 (b) 0.90 (c) 0.73 (d) 0.20 (e) none of the preceding

Solution to Q1: X ∼ B(20, 0.09). So P (X ≤ 2) = 20 0 (.09)0 (.91)20 + 20 1 (.09)1 (.91)19 + 20 2 (.09)2 (.91)18 = 0.73

Q2. A scientist inoculates several mice, one at a time, with a disease germ until he finds one that has contracted the disease. If the probability of contracting the disease is 1/6, what is the probability that 8 mice are required? (The numbers are rounded down to the fourth decimal place). (a) 0.0465 (b) 0.9350 (c) 0.0740 (d) 0.2600 (e) none of the preceding

Solution to Q2: X = # number of mice required to observe the first one with the disease. X ∼ is geometric with p = 1/6. So P (X = 8) = (1 − p)7 p = 0.0465

Q3. Assume that random variables X and Y are independent and have distribution X ∼ Poisson(2) and Y ∼ B(5, 0.2), respectively. Compute P ({X = 0} or {Y = 0}) = P ({X = 0} ∪ {Y = 0}) . (The numbers are rounded up to the fourth decimal place) (a) (d) 0.4325 0.4187 (b) (e) 0.5621 (c) 0.4630 insufficient information provided

Solution to Q3: P ({X = 0} or {Y = 0}) = P ({X = 0}) + P ({Y = 0}) − P ({X = 0} and {Y = 0}) = P ({X = 0}) + P ({Y = 0}) − P ({X = 0}) P ({Y = 0}) = exp(−2) + 0.20 ∗ 0.85 − exp(−2) ∗ 0.20 ∗ 0.85 = 0.4187

Q4. 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 Canada 7.7% of the population aged 65 and over have Alzheimer’s disease.

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Find the probability that a person has the disease given that the test was positive. (The numbers are rounded down to the second decimal place). (a) 0.97 (b) 0.88 (c) 0.99 (d) 0.12 (e) none of the preceding

Solution to Q4: A - test positive, D - a person has disease. Given: P (A|D) = P (Dc |A) (Bayes’ formula): P (D|A) = 436 5 , P (A|Dc ) = , P (D) = 0.077. To find: 450 500

P (A|D)P (D) = 0.88. P (A|D)P (D) + P (A|Dc )P (Dc )

Q5. Consider the following system with four components. We say that it is functional if there exists a path of functional components from left to right. The probability of each component functions is shown. Assume that the components function or fail independently. What is the probability that the system operates? (The numbers are rounded down to the second decimal place). 2 0.7 1 0.5 3 0.4

4 0.5 (a) 0.32 (b) 0.16 (c) 0.03 (d) 0.68 (e) none of the preceding

Solution to Q5: Call ’Box B’ - components 2,3,4, ’Box C’ - components 2,3. P (Box C operates) = P (component 2 operates and component 3 operates) = P (component 2 operates)P (component 3 operates) = 0.4 × 0.7 = 0.28. P (Box B operates) = P (Box C operates or component 4 operates) = P (Box C operates) + P (component 4 operates) − P (Box C operates)P (component 4 operates) = 0.28 + 0.5 − 0.28 ∗ 0.5 = 0.64. P (system operates) = P (component 1 and Box B operate) = P (component 1 operates)P (Box B operates) = 0.5 ∗ 0.64 = 0.32.

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Q6. A material is studied for a...
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