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sc I The discrete random variable X has the following probability mass function: P(X = x) = kx for the values of x = 2,4 and
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x = 2 P[x = x] = ka x = 2,4,5. P[ X = 2] = 2k P[X=4] = 4k P[x=5 ] =sk L x P[x=x Tokel 2 4 1 zk 4k s sk I since it is probabil1 2 _2 4 _1 5 _S_ _PLx-»] ex) - 2XP Cxzx] - 2X2 + 444 + 5 : 4S E) = 14.09 9 9 | 1Ex2) : 2 22 Exzx 2ܡܢ - xf 25 + ܟ * I + 11 ܕDate V(x) = Ecx? - Exit = 17.9091- [4.ogog] = 17.90g - 16.7355 V(x) = 1.1736 F(x) F(x ) E(x)

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