# hw3sol - Engineering 0020 Probability Statistics for...

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Engineering 0020: Probability & Statistics for Engineers 1 Spring 2007 Homework #3: Solutions 2. X = 1 if a randomly selected computer memory module is functional and X = 0 otherwise. X = 1 if a randomly selected automobile has automatic transmission and X = 0 otherwise. X = 1 if a randomly selected item is biodegradable and X = 0 otherwise. X = 1 if a randomly selected chemical sample has pH below 5.5 and X = 0 otherwise. etc. etc. etc. 14. a = 5 1 ) ( y y p = k+2k+3k+4k+5k = 15k. But = 5 1 ) ( y y p = 1 k = 1/15 b P(Y 3) = p (1) + p (2) + p (3) = 1/15+2/15+3/15 = 6/15= 0.4 c P(2 Y 4) = p (2) + p (3) + p (4) = 2/15+3/15+4/15 = 9/15 = 0.6 d NO - because () = 5 1 2 50 y y = 1/50 [1 + 4 + 9 + 16 + 25] = 55/50 1 31. a E(X) = (13.5)(.2) + (15.9)(.5) + (19.1)(.3) = 16.38 E(X 2 ) = (13.5) 2 (.2) + (15.9) 2 (.5) + (19.1) 2 (.3) = 272.298 V(X) = E(X 2 ) - [E(X)] 2 = 272.298 - (16.38) 2 = 3.9936 b E[25X-8.5] = 25E[X] - E[8.5] = (25) (16.38) - 8.5 = 401 c V[25X-8.5] = V[25X] = (25) 2 V(X) +V(8.5) = (625)(93.9939) + 0 = 2496 d E[ h (X)] = E[X-0.01X 2 ] = E[X] - 0.01E[X 2 ] = 16.38 - 2.72 = 13.66 Alternatively, you could do this as E[ h (X)] = Σ x h (x)* p (x) = (13.5-.01*13.5 2 )(0.2) + (15.9-.01*15.9 2 ) (0.5) + (19.1-.01*19.1 2 )(0.3) = 13.66 36. E[X] = + = = = = = = 2 ) 1 ( 1 1 1 ) ( 1 1 1 n n n x n n x x xp n x n x n x = ( n +1)/2 To find V(X) we also need E[X 2 ]:

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## This homework help was uploaded on 04/10/2008 for the course ENGR 0020 taught by Professor Rajgopal during the Spring '08 term at Pittsburgh.

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hw3sol - Engineering 0020 Probability Statistics for...

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