Test 5 Solution

Test 5 Solution - Stat 230 Quiz #5 November 24, 2004 1. Let...

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Stat 230 Quiz #5 November 24, 2004 1. Let X be a continuous random variable with probability density function (p.d.f.) f ( x ) = kx 2 for 0 < x < 2, otherwise f ( x ) = 0. (a) [ 3 marks ] Show that k = 3 / 8. Soln : 1 = R + -∞ f ( x ) dx = k R 2 0 x 2 dx = 8 k/ 3, so k = 3 / 8. (b) [ 3 marks ] Find the cumulative distribution function (c.d.f.) F ( x ) of X . Soln : F ( x ) = 0 if x 0 and F ( x ) = 1 if x 2; for x (0 , 2), F ( x ) = Z x 0 3 u 2 8 du = x 3 8 . (c) [ 3 marks ] Find P (1 < X < 3). Soln : P (1 < X < 3) = F (3) - F (1) = 1 - 1 / 8 = 7 / 8. (or P (1 < X < 3) = R 3 1 f ( x ) dx = R 2 1 (3 x 2 / 8) dx = 7 / 8) (d) [ 3 marks ] Find V ar ( X ). Soln : E ( X ) = R 2 0 x (3 x 2 / 8) dx = 3 × 2 4 / (4 × 8) = 3 / 2; E ( X 2 ) = R 2 0 x 2 (3 x 2 / 8) dx = 3 × 2 5 / (5 × 8) = 12 / 5. So V ar ( X ) = 12 / 5 - (3 / 2) 2 = 3 / 20 = 0 . 15. (e) [ 3 marks ] Let Y = ln( X ). Find the p.d.f. of Y . Soln
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This note was uploaded on 10/21/2010 for the course STAT 230 taught by Professor Various during the Fall '06 term at Waterloo.

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Test 5 Solution - Stat 230 Quiz #5 November 24, 2004 1. Let...

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