Test 3 Solutions

Test 3 Solutions - STAT 230 Test 3 3:30 4:10 pm 1 A...

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STAT 230 Test 3 June 14, 2006 3:30 – 4:10 pm 1. A hardware company conducts tests on computer graphic cards. These cards are made of 10 components which work independently. Each component has probability 0.03 of being defective. A graphic card is defective if at least two of the components are defective. Each test on one card takes exactly two minutes and tests are conducted one immediately after another. [2] (a) Show that the probability of a graphic card being defective is 0.035 Let X be the number of defective components on a card, then X Binomial( n = 10 ,p = 0 . 03). P (card works) = P ( X = 0) + P ( X = 1) = (1 - 0 . 03) 1 0 + 10 × 0 . 03 × (1 - 0 . 03) 9 = 0 . 9654 P (card defective) = 1 - P (card works) = 1 - 0 . 9654 = 0 . 035. [2] (b) Give an expression for the probability that among 80 tested cards, there are exactly 4 defective cards. Let X be the number of defective cards among the 80 cards, then X Binomial( n = 80 ,p = 0 . 035). P ( X = 4) = ± 80 4 ² 0 . 035 4 × (1 - 0 . 035) 76 (= 0 . 158). [2] (c) Use a suitable approximation to calculate the probability in (b). X can be approximated by Poisson( μ = 80 × 0 . 035 = 2 . 8): P ( X = 4) . = e - 2 . 8 × 2 . 8 4 / 4! = 0 . 156. [2] (d) Find the probability that, once tests start, we will have to wait for exactly 20 minutes to get a defective card. Let X be the number of non-defective cards tested before observing a defective one. Then X Geo( p = 0 . 035). The 20 minutes waiting time corresponds to a total number of 10 cards tested (9 non-defective and 1 defective). P ( X = 9) = (1 - 0 . 035) 9 × 0 . 035 = 0 . 025. [2] (e) Suppose the goal is to find 2 defective cards, and this goal has not been achieved after 20 minutes from the start of the tests. Give an expression for the probability that we will have to wait for at least 10 more minutes to get what we want. Let
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Test 3 Solutions - STAT 230 Test 3 3:30 4:10 pm 1 A...

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