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practicefinalsolutions - STAT 400 Final Exam Practice...

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STAT 400 Final Exam Practice Problems 1. A chain of video stores sells three different brands of VCRs. Of it’s VCR sales, 60% are brand 1 (the least expensive), 30% are brand 2, and 10% are brand 3. Each manufacturer offers a 1-year 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 · 0 . 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 · 0 . 6 + 0 . 2 · 0 . 3 + 0 . 1 · 0 . 1 = 0 . 22 (c) P ( A 1 | B ) = P ( B A 1 ) P ( B ) = 0 . 15 0 . 22 = 0 . 682. 1
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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, π 0 . 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 ˆ 16 X i =1 X 2 i < c ! = 0 . 95 Ans: X N (470 , 100 2 ). (a) P ( X π 0 . 33 ) = 0 . 33, i.e. P ( X - 470 100 π 0 . 33 - 470 100 ) = P ( Z π 0 . 33 - 470 100 ) = 0 . 33, so by symmetry property, P ( Z 470 - π 0 . 33 100 ) = 0 . 67. From the normal table a, we get 470 - π 0 . 33 100 = 0 . 44. Therefore, π 0 . 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 ( - 0 . 63 < Z < 2 . 53) = Φ(2 . 53) - (1 - Φ(0 . 63)) = 0 . 9943 - 0 . 2643 = 0 . 73 (c) 16 i =1 X 2 i χ 2 (16), so c = χ 2 0 . 05 (16) = 26 . 30. 2
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3. Let X equal the fill weight in April and Y the fill weight in June for an 8-pound box of bleach. We shall test the null hypothesis H 0 : μ Y - μ X = 0 against the alternative hypothesis H 1 : μ X - μ Y > 0 given that n = 90 observations of X yielded ¯ x = 8 . 10, s x = 0 . 117 and m = 110 observations of Y yielded ¯ y = 8 . 07 and s y = 0 . 054.
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