Unformatted text preview: MAT 2378 Midterm practice 26 October, 2009 Time: 70 minutes Student Number: Family Name: First Name: Professor: R. Kulik This is a closed book examination. Only nonprogrammable and nongraphic calculators are permitted. Record your answer to each question in the table below. Number of pages: . 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 Answer 1 1 Q1. (2.23) As a part of a classic experiment on mutations, ten aliquots of identical size were taken from the same culture of the bacterium E.coli. For each aliquot, the number of bacteria resistant to a certain virus was determined. The results were as follows: 14 15 13 21 14 14 26 16 20 13 The mean, median and the third quartile are, respectively, : (a) (d) 16.6; 14.5; 14 18.1; 14.5; 20 (b) (e) 16.6; 14.5; 20 none of the preceding (c) 16.6; 14; 20 Q2. Suppose a certain drug test is 99% sensitive and 99% speciﬁc, that is, the test will correctly identify a drug user as testing positive 99% of the time, and will correctly identify a nonuser as testing negative 99% of the time. Let’s assume a corporation decides to test its employees for opium use, and it is known 0.5% of the employees use the drug. The probability that, given a positive drug test, an employee is actually a drug user, is: (a) 0.3322 (b) 0.1622 (c) HINT: compare Question 1 in Assignment 3. Solution to Q2: Let D be the event of being a drug user and N indicate being a nonuser. Let A be the event of a positive drug test. To compute: P (DA). We know the following: • P (D), or the probability that the employee is a drug user. This is 0.005. • P (N ), or the probability that the employee is not a drug user. This is 1 − P (D), or 0.995. • P (AD), or the probability that the test is positive, given that the employee is a drug user. This is 0.99, since the test is 99% accurate. • P (AN ), or the probability that the test is positive, given that the employee is not a drug user. This is 0.01, since the test will produce a false positive for 1% of nonusers. Compute ﬁrst P (A) = P (AD)P (D) + P (AN )P (N ) and P (A and D). Once you have this, compute P (DA) = P (A and D) = 0.3322 P (A) 0.99 (d) 0.01 (e) none of the preceding Q3. A city has 1000 married couples with both husband and wife working. Each person was asked whether his or her salary exceeded $30,000. The following information was obtained. Wife Wife less than $30,000 more than $30,000 Husband less than $30,000 430 60 Husband more than $30,000 410 100 What is the probability that a wife earns more than $30,000 given that the husband earns more than $30,000? (a) 0.8059 (b) 0.1961 (c) 0.4300 (d) 0.5700 (e) none of the preceding 2 Solution to Q3: Let A be the event that a wife earns more than $30,000, B be the event that the husband earns more than $30,000. To calculate P (AB ). We have P (A and B ) = 100/1000 = 0.1, P (B ) = 510/1000 = 0.51 P (AB ) = P (A and B )/P (B ) = 0.1/0.51 = 0.1961 . Q4. Let Y be a discrete random variable with probability distribution function given below. y 0 1 2 P (Y = y ) 0.5 0.3 0.2 The expected value and the variance of Y are, respectively: (a) 0.7, 0.61 (b) 0.7, 1.1 (c) 0.5, 0.61 (d) 0.5, 1.1 (e) none of the preceding. 3 Q5. In a group of twenty students, each student has a probability of 0.7 of passing an exam. What is the probability that more than 11 of them will pass an exam? (a) 0.9829 (b) 0.0171 (c) 0.0480 (d) 0.8866 (e) none of the preceding. You may use the following MINITAB output: Binomial with n = 20 and p = 0.7 x 0 1 2 3 4 5 6 7 8 9 10 11 P( X = x ) 0.0000000 0.0000000 0.0000000 0.0000005 0.0000050 0.0000374 0.0002181 0.0010178 0.0038593 0.0120067 0.0308171 0.065370 Solution to Q5: X  number of students who pass an exam. X ∼ B (20, 0.7). To compute P (X > 11) = 1 − P (X ≤ 11) = 1 − 0.11333 = 0.88667. Q6. A company manufactures hockey pucks. It is known that their weight is normally distributed with mean 1 and the standard deviation 0.05. The pucks used by NHL must weight between 0.9 and 1.1. What is the percentage of pucks produced by the company, which could be used by NHL? (a) 100% (b) 95.45% (c) 4.56% (d) 97.72% (e) none of the preceding Solution to Q6: Let X  weight of a puck. Then P (0.9 < X < 1.1) = P ( 0.9 − 1.0 1.1 − 1.0 <Z< ) = P (Z < 2) − P (Z < −2) = 0.977250 − 0.022750 = 0.9545 0.05 0.05 Q7. (5.16) Forced expiratory volume (FEV) is the volume of air that a person can expire in one second. A physician takes a random sample of n = 15 young women from a normal population with mean µ = 3000mLi and standard deviation σ = 400mLi. The probability that the sample mean will be within ±100mLi of the true population mean, is (a) 0.9476 (b) 0.3222 (c) 0.1463 (d) 0.6680 (e) none of the preceding Solution to Q7: 4 2 Y1 , . . . , Yn  average FEV of each person, Yi ∼ N (3000, 4002 ). We have µY = 3000, σY = 4002 /15, so that ¯ ¯ σY = 103.3. To compute ¯ ¯ 2900 − 3000 Y − 3000 3100 − 3000 ¯ P (2900 ≤ Y ≤ 3100) = P ≤ ≤ = P (−0.97 ≤ Z ≤ 0.97) = 0.6680 103.3 103.3 103.3 Q8. Assume that we have a sample X1 , . . . , X100 from an arbitrary population with mean µ = 1 and variance σ 2 = 4. The approximate probability that the sample mean exceeds 1.5, is: (a) 0.99379 (b) 0.00621 (c) 0.5000 (d) 0.7742 (e) none of the preceding Solution to Q8: ¯ P (X > 1.5) = P ¯ X −µ 1.5 − 1 > σ 2 /n 4/100 = P (Z > 2.5) = 1 − 0.993790 = 0.00621 Q9. The examination scores in an university course have mean 56 and sd 11. In a class of 100 students, what is the approximate probability that the average mark lies between 54 and 58? (a) 0.9309 (b) 0.4648 (c) 0.8622 (d) 0.2465 (e) none of the preceding Solution to Q9: ¯ Solution: Let the marks be Y1 , .., Y49 . The average is Y ∼ N (56, 112 /100). Then ¯ < 58) ≈ P − 54 − 56 < Z < 58 − 56 P (54 < Y 11/10 11/10 = P (−1.8181 < Z < 1.8181) = 0.9309. 5 Q10. Which of the following set of ﬁgures describes data coming from the same normal population with mean 5? (a) (b) (c) (d) This is the last question 6 Solutions to multiple choice questions: Q1 −→ b Q2 −→ a Q4 −→ a Q3 −→ b Q5 −→ d Q6 −→ b Q7 −→ d Q8 −→ b Q10 −→ c Q9 −→ a ...
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 Spring '09
 Kulik
 Probability, drug user

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