ma316f03hw5

# ma316f03hw5 - MA 316 MATHEMATICAL PROBABILITY ASSIGNMENT#5...

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MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #5 SOLUTIONS: SECTIONS 3.1 and 3.2 FALL 2003 3.3: If X B ( n,p ) then E [ X ] = μ = np and V AR [ X ] = σ 2 = np (1 - p ). Using that expectation is a linear function we have E [ X/n ] = (1 /n ) E [ X ] = ( np ) /n = p and E [( X/n - p ) 2 ] = 1 /n 2 E [( X - pn ) 2 ] = 1 /n 2 E [( X - μ ) 2 ] = (1 /n 2 ) σ 2 = ( np (1 - p )) /n 2 = p (1 - p ) /n. 3.6: If X ( n, 1 / 4), to determine the smallest n such that P (1 Y ) 0 . 70 we can ﬁnd the smallest integer n for which P (1 Y ) = 1 - P ( Y = 0) = 1 - ± 3 4 n ± 1 4 0 0 . 7 or ( 3 4 ) n 0 . 3 so n = 5 . 3.13: If f ( x ) = ( 1 3 )( 2 3 ) x for x = 0 , 1 , 2 ,... then the conditional density function that X = x given that X 3 is P ( X = x | X 3) = P (( X = x ) ( X 3)) P ( X 3) = P ( X = x ) P ( X 3) = 1 / 3(2 / 3) x 1 / 3 x =3 (2 / 3) x . 3.15: Using that the Maclaurin series for (1 - w ) r is (1 - w ) r = X y =0 ± y + r - 1 r - 1 w y = 1 + rw + r ( r

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ma316f03hw5 - MA 316 MATHEMATICAL PROBABILITY ASSIGNMENT#5...

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