Review_Problems-Solutions

# Review_Problems-Solutions - Final review Practice problems...

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Unformatted text preview: Final review: Practice problems - SOLUTIONS Note: These are only outlines of the solutions. You have to show all work on the test to get full credit. 1. P ( delivered on time | ready on time ) = . 72 /. 80 = 0 . 9 2. S = strike, T = completed on time, P ( S ) = 0 . 6, P ( T | S ) = 0 . 65, P ( T | S ) = 0 . 85 P ( T ) = P ( T ∩ S ) + P ( T ∩ S ) = ( . 6)( . 35) + ( . 4)( . 85) = 0 . 55 P ( S | T ) = P ( T ∩ S ) P ( T ) = ( . 6)( . 35) . 55 = 0 . 382 3. 78- (64+36- 34) 78 = 12 78 4. S = survive operation, R = reject transplant, P ( S ) = 0 . 55, P ( R | S ) = 0 . 2 P ( survive both stages ) = P ( S ∩ R ) = ( . 8)( . 55) = . 44 5. P ( at least one) = 1- P ( none ) = 1- (1- . 10) 1 2 = 1- (0 . 9) 1 2 = 0 . 718 6. If X had the distribution function ... (a) Discrete (b) ∑ 5 x =0 p ( x ) = 1 ⇒ c = 1 / 75 (c) We can write the pmf p ( x ) as x 1 2 3 4 5 p ( x ) 1/15 2/15 3/15 4/15 5/15 F ( x ) = , for x < 1 1 / 15 , for 1 ≤ x < 2 3 / 15 , for 2 ≤ x < 3 6 / 15 , for 3 ≤ x < 4 10 / 15 , for 4 ≤ x < 5 1 , x ≥ 5 (d) E ( X ) = 3 . 67, V ar ( X ) = 15- (3 . 67) 2 = 1 . 56 7. If X had the distribution function... (a) Discrete (b) P (2 < X ≤ 6) = 5 / 6- 1 / 3 = 3 / 6 = 1 / 2 (c) P ( X = 4) = 1 / 2- 1 / 3 = 1 / 6 (d) We can write the pmf p ( x ) as...
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Review_Problems-Solutions - Final review Practice problems...

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