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Unformatted text preview: . 1718 . Problem 4 Given that X Poi ( = 6), we get P [ X 4] = 4 X i =0 P [ X = i ] = 4 X i =0 e6 6 i i ! = 0 . 2851 . Using normal approximation, we get X = X6 6 N (0 , 1) . So, P [ X 4] = P X6 6 2 6 = 2 6 = 0 . 2071 . We know that the normal approximation is valid for large values of . So, even though here = 6 is not very large, the approximated probability is close (though certainly not very close) to the actual one. Problem 5 Let F ( ) be the cumulative distribution function of Y = 1 b Xa b . Then, F ( x ) = P 1 b Xa b x = P [ X bx + a ] = bx + aa b = ( x ) . So, the distribution of Y is standard normal. So, X = bY + a N ( a,b 2 ) . 2...
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