MAT 2377C Final Exam
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
1
2
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) …show more content…
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.
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
1
2
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) …show more content…
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.