chapter5 - 1 Chapter 5 51. The probability mass function is...

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1 Chapter 5 5–1. The probability mass function is x p ( x ) 0 (1 - p ) 4 1 4 p (1 - p ) 3 2 6 p 2 (1 - p ) 2 3 4 p 3 (1 - p ) 4 p 4 otherwise 0 5–2. P ( X 5) = 6 X x =5 ± 6 x (0 . 95) x (0 . 05) 6 - x = 6(0 . 95) 5 (0 . 05) + (0 . 95) 6 = 0 . 9672 5–3. Assume independence and let W represent the number of orders received. P ( W 4) = 12 X w =4 ± 12 w (0 . 5) w (0 . 5) 12 - w = (0 . 5) 12 12 X w =4 ± 12 w = 1 - (0 . 5) 12 3 X w =0 ± 12 w = 0 . 9270 5–4. Assume customer decisions are independent. P ( X 10) = 1 - P ( X 9) = 1 - 9 X x =0 ± 20 x ¶± 1 3 x ± 2 3 20 - x = 0 . 0918 5–5. P ( X > 2) = 1 - P ( X 2) = 1 - 2 X x =0 ± 50 x (0 . 02) x (0 . 98) 50 - x = 0 . 0784 .
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2 5–6. M X ( t ) = E [ e tX ] = n X x =0 e tx ± n x p x (1 - p ) n - x = ( pe t + q ) n , where q = 1 - p E [ X ] = M 0 X (0) = [ n ( pe t + q ) n - 1 pe t ] t =0 = np E [ X 2 ] = M 00 X (0) = np [ e t ( n - 1)( pe t + q ) n - 2 ( pe t ) + ( pe t + q ) n - 1 e t ] t =0 = ( np ) 2 - np 2 + np V ( X ) = E [ X 2 ] - ( E [ X ]) 2 = n 2 p 2 - np 2 + np - n 2 p 2 = np (1 - p ) = npq 5–7. P p 0 . 03) = P ± X 100 0 . 03 = P ( X 3) = 3 X x =0 ± 100 3 (0 . 01) x (0 . 99) 100 - x = 0 . 9816 5–8. P p > p + p pq/n ) = P ± ˆ p > 0 . 07 + p (0 . 07)(0 . 93) / 200 = P ( X > 200(0 . 088)) = P ( X > 17 . 6) = 1 - P ( X 17 . 6) = 1 - 17 X x =0 ± 200 x (0 . 07) x (0 . 93) 200 - x = 0 . 1649
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3 By two standard deviations, P p > p + 2 p pq/n ) = P p > 0 . 106) = P ( X > 21 . 2) = 1 - P ( X 21) = 1 - 21 X x =0 ± 200 x (0 . 07) x (0 . 93) 200 - x = 0 . 0242 By three standard deviations, P p > p + 3 p pq/n ) = P ( X > 24 . 8) = 1 - P ( X 24) = 1 - 24 X x =0 ± 200 x (0 . 07) x (0 . 93) 200 - x = 0 . 0036 5–9. P ( X = 5) = (0 . 95) 4 (0 . 05) = 0 . 0407 5–10. A : Successful on first three calls, B : Unsuccessful on fourth call P ( B | A ) = P ( B ) = 0 . 90 if A and B are independent 5–11. P ( X = 5) = p 4 (1 - p ) = f ( p ) df ( p ) dp = 5 p 4 - 4 p 3 = 0 p = 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 f(p) p
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4 5–12. (a) X = trials required up to and including the first sale C ( X ) = 1000 X + 3000( X - 1) = 4000 X - 3000 E [ C ( X )] = 4000 E [ X ] -
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue University-West Lafayette.

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chapter5 - 1 Chapter 5 51. The probability mass function is...

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