Exam 3 Comments

Exam 3 Comments - Math 351 Test 3 Some comments on the test...

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October 31, 2010 Math 351 - Test 3 Some comments on the test. 1. Problem 1: (a) Most correctly deduced that p = 1 / 3. You should view it as an urn problem. You can then calculate p either by using combinations, or by conditional probability (it is the probability the first is black and the second is black). (b) You should recognize in a given word problem when a binomial, geometric or negative binomial random variable is being described. All three are discrete and arise when we have repeated inde- pendent trials of an experiment in which on each trial we can only have success (with probability p ) or failure. Call our random variable Z . Then Z is binomial with parameters n and p if Z is the number of successes in n trials. Z is geometric with parameter p if Z is the number of trials until the first success. Z is negative binomial with parameters p and r if Z is the number of trials to the rth success. Thus in the given problem, X is geometric with parameter p = 1 / 3 and Y is negative binomial with parameters p = 1 / 3 and r = 5. It’s true that a geometric rv with parameter p = 1 / 3 can also be viewed as a negative binomial with parameter p = 1 / 3 and r = 1, but it’s best to identify it as geometric. (c) You just need to remember that E [ X ] = 1 /p . If I had asked instead for E [ Y ], the answer would have been r/p = 5 / (1 / 3) = 15 (d) It’s hard to remember the formula for the probability mass function of a negative binomial, so it’s best to recall where it comes from. To find P { Y = 10 } , the 5th success must come on the 10th trial, so this means we have exactly 4 successes in the first 9 trials (calculated using the formula for a binomial random variable with parameters n = 9 and p = 1 / 3 equaling 4) and we have a success on the 10th trial. 2. Problem 2: The first step is to write the random variable Y in terms of X : Y = 0 if X < 0 or X > 15 12 if 0 < X < 12 X - 12 if 12 < X < 15 Thus we’ve managed to write Y in the form Y = g ( X ), where g is a real-valued function from R to R , namely g ( x ) = 0 if x < 0 or x > 15 12 if 0 < x < 12 x - 12 if 12 < x < 15 The key is to make use of the formula E [ Y ] = E [ g ( X )] = Z -∞ g ( x ) f X ( x ) dx. This is preferable to using the formula E [ Y ] = Z -∞ y f Y ( y ) dy since it’s not that obvious what is f Y ( y ) (see comments on Problem 4).
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This note was uploaded on 03/05/2011 for the course MATH 351 taught by Professor Moumen,f during the Spring '08 term at George Mason.

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Exam 3 Comments - Math 351 Test 3 Some comments on the test...

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