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Unformatted text preview: STAT 400 Final Exam Practice Problems 1. A chain of video stores sells three different brands of VCRs. Of its VCR sales, 60% are brand 1 (the least expensive), 30% are brand 2, and 10% are brand 3. Each manufacturer offers a 1year warranty on parts and labor. It is known that 25% of brand 1 VCRs require warranty work, whereas the corresponding percentages for brands 2 and 3 are 20%, 10%, respectively. (a) What is the probability that a randomly selected purchaser has bought a brand 1 VCR and the VCR will need repair while under warranty? (b) What is the probability that a randomly selected purchaser has a VCR that will need repair under warranty? (c) If a customer returns to the store with a VCR that needs warranty repair work, what is the probability that it is a brand 1 VCR? Ans: Let B = event VCR purchased need warranty repair work, A i = event VCR purchased is brand i , i = 1 , 2 , 3. Information given: P ( A 1 ) = 0 . 6, P ( A 2 ) = 0 . 3, P ( A 3 ) = 0 . 1, P ( B  A 1 ) = 0 . 25, P ( B  A 2 ) = 0 . 2 and P ( B  A 3 ) = 0 . 1. (a) P ( A 1 B ) = P ( B  A 1 ) P ( A 1 ) = 0 . 25 . 6 = 0 . 15. (b) By the partition theorem, P ( B ) = P ( B  A 1 ) P ( A 1 ) + P ( B  A 2 ) P ( A 2 ) + P ( B  A 3 ) P ( A 3 ) = 0 . 25 . 6 + 0 . 2 . 3 + 0 . 1 . 1 = 0 . 22 (c) P ( A 1  B ) = P ( B A 1 ) P ( B ) = . 15 . 22 = 0 . 682. 1 2. (a) Suppose that math SAT scores for a particular population are normally distributed with mean 470 and a standard deviation of 100. i. Find the 33rd percentile, . 33 . ii. Given a random sample of size n = 10 scores, find P (450 < X < 550). (b) Suppose X 1 ,X 2 , ,X 16 is a random sample of size 16 from N (0 , 1). Determine c , such that P 16 X i =1 X 2 i < c ! = 0 . 95 Ans: X N (470 , 100 2 ). (a) P ( X . 33 ) = 0 . 33, i.e. P ( X 470 100 . 33 470 100 ) = P ( Z . 33 470 100 ) = 0 . 33, so by symmetry property, P ( Z 470 . 33 100 ) = 0 . 67. From the normal table a, we get 470 . 33 100 = 0 . 44. Therefore, . 33 = 426. (b) X N (470 , 100 2 / 10) = N (470 , 1000) From the normal table a, we get P (450 < X < 550) = P 450 470 1000 < X 470 1000 < 550 470 1000 = P ( . 63 < Z < 2 . 53) = (2 . 53) (1 (0 . 63)) = 0 . 9943 . 2643 = 0 . 73 (c) 16 i =1 X 2 i 2 (16), so c = 2 . 05 (16) = 26 . 30....
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This note was uploaded on 09/11/2011 for the course STAT 400 taught by Professor Tba during the Spring '05 term at University of Illinois at Urbana–Champaign.
 Spring '05
 TBA
 Statistics, Probability

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