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Unformatted text preview: Engineering Statistics ENGRD2700 SOLUTIONS PRACTICE FINAL EXAM Spring 2009 1. (a) Pr[4 or 6 in 3 draws] = Pr[4 or 6 in 1 draw] 3 = ( 1 5 ) 3 = 1 125 = 0 . 008 (b) Test H : p = 0 . 008 versus H 1 : p > . 008. Onesided is most appropriate. ˆ p = 15 1200 = 0 . 0125 . n = 1200 is big so can use normal approximation. (Poisson approx is also OK.) Z = . 0125 . 008 p ( . 008)( . 992) / 1200 = 1 . 75 So P = . 04. If we use a onesided test with α = 0 . 05 (cutoff is 1.645), then we may reject H in favor of H 1 and conclude the lottery was fixed. 2. Let the event D be { Drug user } and event FF be { failed two tests } . Pr[ D  FF ] = Pr[ FF  D ] · Pr[ D ] Pr[ FF  D ] · Pr[ D ] + Pr[ FF  D ] · Pr[ D ] = ( . 98)( . 98)( . 05) ( . 98)( . 98)( . 05) + ( . 03)( . 03)( . 95) = . 9825 . 3. (a) N is geometric. E ( N ) = 1 /p = 5. (b) Pr[ N = 3] = ( . 8) 2 ( . 2) = . 128. (c) Pr[2 out of 6] = b (2; n = 6 ,p = . 2) = 6 2 ( . 8) 4 ( . 2) 2 = 0 . 24576. 4. Let X be the pin length. Let S , E be the events that the pin is made by stamping and extrusion processes respectively. (a) The required probability is given by: P (0 . 5 ≤ X ≤ . 9) = P (0 . 5 ≤ X ≤ . 9  S ) P ( S ) + P (0 . 5 ≤ X ≤ . 9  E ) P ( E ) = 0 . 3 × Z . 9 . 5 3 x 2 dx + 0 . 7 × Z . 9 . 5 2 xdx = 0 . 5732 (b) We use the Bayes rule and calculations from the previous part of the question. P ( E  . 5 ≤ X ≤ . 9) = P (0 . 5 ≤ X ≤ . 9  E ) P ( E ) P (0 . 5 ≤ X ≤ . 9) = . 392 . 5732 = 0 . 684 (c) The required expected strength can be computed as E 2 X = Z 1 2 x (0 . 3 × f s ( x ) + 0 . 7 × f e ( x )) dx = 2[0 . 3 × 3 2 + 0 . 7 × 2] = 3 . 7 1 5. Let X be the pin length. Let S , E be the events that the pin is made by stamping and extrusion processes, respectively. Let D be the event that the pin is defective....
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This note was uploaded on 09/06/2009 for the course ENGRD 2700 taught by Professor Staff during the Spring '05 term at Cornell.
 Spring '05
 STAFF

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