hw5solu_427

# hw5solu_427 - HW5 Solutions 2 f(x = a b c d 4 a b 1 10...

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HW5: Solutions Chapter 4: 2. f(x) = 10 1 for –5 x 5, and = 0 otherwise a. P(X < 0) = 5 . 0 5 10 1 dx b. P(–2.5 < X < 2.5) = 5 . 5 . 2 5 . 2 10 1 dx c. P(–2 X 3) = 5 . 3 2 10 1 dx d. P(k < X < k + 4) = 4 . ] ) 4 [( 10 1 4 10 4 10 1 k k dx k k x k k 4. a. 1 ) 1 ( 0 ) ; ( 0 2 / 0 2 / 2 2 2 2 2 x x e dx e x dx x f b. P(X 200) = 200 0 2 / 2 200 2 2 ) ; ( dx e x dx x f x 8647 . 1 1353 . 200 0 2 / 2 2 x e P(X < 200) = P(X 200) .8647, since x is continuous. P(X 200) = 1 – P(X < 200) .1353 c. P(100 X 200) = 200 100 ) ; ( dx x f 4712 . 200 100 000 , 20 / 2 x e d. For x > 0, P(X x) = x dy y f ) ; ( x y dx e e y 0 2 / 2 2 2 2 2 2 2 2 / 0 2 / 1 x x y e e 6. a. b. 1 = 1 1 2 4 2 2 4 3 3 4 ] 1 [ ] ) 3 ( 1 [ k du u k dx x k c. P(X > 3) = 5 . ] ) 3 ( 1 [ 4 3 2 4 3 dx

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## This note was uploaded on 03/22/2011 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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hw5solu_427 - HW5 Solutions 2 f(x = a b c d 4 a b 1 10...

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