3 20 marks a continuous random variable also

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Unformatted text preview: cumulative distribution function, which is 0, , 0 1, (a) (b) (c) (d) [4’] Find out the coefficient ; [4’] Find the probability density function of ; [4’] Find the probability that 0.1, 0.4 ; [4’] Given we know that 0.5, find the probability that 4 0 1 1 0.1, 0.4 ; (e) [4’] Find the expected value and the variance of Solution: (a) Since is continuous, quantity, therefore 1 (b) By differentiating 1 1 4’ ; 1 . 1, where , we can obtain the following probability density function, 2 1, 0 1 , (c) 0.1 (d) Since 0.1 0.5 0.4| 1 (e) 0.16 0.25, therefore 0.1 0.5 1 2 0.5 0.01 1 0.4 1 0.5 1 0.15 0.1...
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This document was uploaded on 03/19/2014.

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