X z α 2 s n 20 4 032 s 5 2 6 15 and yes that square

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X ± z α/ 2 s n 20 ± (4 . 032) s 5 . 2 6 (15 . 037 , 23 . 101) (...and, yes that square root is correct since I plugged in s 2 instead of s !) 11. n = 1500, ˆ p = 357 / 1500 = 0 . 238 The 95% upper confidence bound for μ is given (approximately) by ˆ p + z α s ˆ p (1 - ˆ p ) n = 0 . 238 + 1 . 645 s (0 . 238)(0 . 762) 1500 0 . 2561
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12. a. To find c we note that the pdf must integrate to 1. Given the constraints, the region where this pdf is non-zero is the lower left triangle half of the unit square. We may set up the integral as c Z 1 0 Z 1 - x 0 (1 - y ) dy dx = 1 or c Z 1 0 Z 1 - y 0 (1 - y ) dx dy = 1 In either case the integral on the left hand side is equal to 1 3 c , so we have 1 3 c = 1 c = 3 . b. We want to integrate out the y -component. For any fixed x , y goes from 0 to 1 - x . So, f X ( x ) = 3 Z 1 - x 0 (1 - y ) dy = 3 y - 1 2 y 2 y =1 - x y =0 = 3[(1 - x ) - 1 2 (1 - x ) 2 ] = 3 2 (1 - x 2 ) for 0 < x < 1. The marginal pdf is zero otherwise. c. We want f X | Y ( x | 0 . 35). f X | Y ( x | 0 . 35) = f ( x, 0 . 35) f Y (0 . 35) First we need to find the marginal pdf for Y f Y ( y ) = 3 Z 1 - y 0 (1 - y ) dx = 3 [(1 - y ) x ] x =1 - y x =0 = 3(1 - y ) 2 for 0 < y < 1. (Zero otherwise.) So f X | Y ( x | 0 . 35) = f ( x, 0 . 35) f Y (0 . 35) = 3(1 - 0 . 35) 3(1 - 0 . 35) 2 = 1 0 . 65 = 20 13 Note that this is a constant function! What values of x are allowed? From the constraint x + y < 1 with y = 0 . 35 we see that we need x < 0 . 65. From the original constraints, we also see that x > 0. So the marginal distribution for x is the Uniform(0, 0.65) distribution! ie: f X | Y ( x | 0 . 35) = ( 1 0 . 65 , 0 < x < 0 . 65 0 , otherwise
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d. No. There are two ways to justify this. In the above steps we have found the marginal densities f X ( x ) and f Y ( y ) and we can see that f ( x, y ) 6 = f X ( x ) f Y ( y ) . Also, we can see that f X | Y ( x | y ) 6 = f X ( x ). e. ρ X,Y = Cov ( X, Y ) q V ar ( X ) V ar ( Y ) where Cov ( X, Y ) = E [ XY ] - E [ X ] E [ Y ]. Now E [ X ] = Z -∞ x f X ( x ) dx = 3 2 Z 1 0 x (1 - x 2 ) dx = 3 8 E [ X 2 ] = Z -∞ x 2 f X ( x ) dx = 3 2 Z 1 0 x 2 (1 - x 2 ) dx = 1 5 V ar [ X ] = E [ X 2 ] - ( E [ X ]) 2 = 1 5 - 3 8 2 = 19 320 and E [ Y ] = Z -∞ y f Y ( y ) dx = 3 Z 1 0 y (1 - y ) 2 dx =
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