Statistics - Midterm Practice

Statistics - Midterm Practice - 1 Q1. (2.23) As a part of a...

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1 Q 1. (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) 16.6; 14.5; 14 (b) 16.6; 14.5; 20 (c) 16.6; 14; 20 (d) 18.1; 14.5; 20 (e) none of the preceding Q 2. Suppose a certain drug test is 99% sensitive and 99% speci±c, that is, the test will correctly identify a drug user as testing positive 99% of the time, and will correctly identify a non-user 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) 0 . 99 (d) 0 . 01 (e) none of the preceding 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 non-user. Let A be the event of a positive drug test. To compute: P ( D | A ). 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
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Statistics - Midterm Practice - 1 Q1. (2.23) As a part of a...

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