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Unformatted text preview: Solutions to Exam Two Review Problems 1. (a) P ( X = x  Y = 1) = P ( X = x,Y =1) P ( Y =1) So, the conditional pdf is given by x 1 2 P ( X = x  Y = 1) . 375 . 0625 . 5625 (b) Cov ( X, Y ) = E [ XY ] E [ X ] E [ Y ] E [ XY ] = (0)(0)(0 . 10) + (1)(0)(0 . 05) + ··· + (2)(1)(0 . 45) = 0 . 95 E [ X ] = (0)(0 . 4) + (1)(0 . 10) + (2)(0 . 50) = 1 . 1 E [ Y ] = (0)(0 . 2) + (1)(0 . 8) = 0 . 8 So, Cov ( X, Y ) = 0 . 95 (1 . 1)(0 . 8) = 0 . 07 2. n = 1350, ˆ p = 635 / 1350 ˆ p ± z α/ 2 s ˆ p (1 ˆ p ) n For a 95% confidence interval, the zcritical value is 1 . 96. So, the answer is 635 1350 ± (1 . 96) v u u t 635 1350 1 635 1350 1350 (0 . 4437 , . 4970) 3. (a) n = 36, x = 1 . 25, s = 0 . 75 “Large” sample ( n > 30) use a zcritical value. X ± z α/ 2 s √ n 1 . 25 ± (1 . 44) . 75 √ 36 (1 . 07 , 1 . 43) (b) For a narrower interval, we must increase the sample size. 4. n = 15, x = 53 . 870, s = 6 . 82 Small sample, “ s instead of σ ” implies a tcritical value with n 1 = 14 degrees of freedom. 53 . 87 ± (2 . 145) 6 . 82 √ 15 = 53 . 87 ± 3 . 777 5. What are we creating a confidence interval for μ if it is given? Well, mostly because it was a typo and shouldn’t have been given. However, I’ll say that it is simply the belief of these scientists but not a fact. n = 86, x = 11 . 8, s = . 8 Large sample ( n > 30) implies a zcritical value. The critical value for a 70% CI is 1 . 04. X ± z α/ 2 s √ n 11 . 8 ± 1 . 04 . 8 √ 86 (11 . 710 , 11 . 890) 6. (a) f X ( x ) = R 1 6 5 ( x 2 + y ) dy = 6 5 ( x 2 y + . 5 y 2 ) 1 = 6 5 ( x 2 + . 5) for 0 ≤ x ≤ 1. f Y = R 1 6 5 ( x 2 + y ) dx = 6 5 ( 1 3 x 3 + yx ) 1 = 6 5 ( 1 3 + y ) for 0 ≤ y ≤ 1. (b) E [ XY ] = R 1 R 1 xy 6 5 ( x 2 + y ) dx dy = 6 5 R 1 R 1 ( x 3 y + xy 2 ) dx dy = 6 5 R 1 1 4 x 4 y + . 5 x 2 y 2 1 dy = 6 5 ( 1 8 y 2 + 1 6 y 3 1 = . 35 (c) f X  Y ( x  y ) = f ( x,y ) f Y ( y ) = 6 5...
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 Normal Distribution, dx, fX Y

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