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Published byDante Harmison Modified about 1 year ago

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x012 P(x)0.20.60.2 A probability distribution is as shown. If it is repeated and the 2 distributions combined then the following would be the table of potential results Value 012 0 1 2 0 2 42 13 3 0.2x0.2 2 1 Prob’ 012 0 1 2 0.6x0.60.2x0.6 0.6x0.2 0.2x0.6 0.2x0.2 0.6x0.2

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Prob’012 00.040.120.04 10.120.360.12 20.040.120.04 y01234 P(y)0.040.240.440.240.04 yP(y)00.240.880.720.16 y 2 P(y)00.241.762.160.64 E(y)= y P(y)] =0+0.24+0.88+0.72+ 0.16 = 2.00 2 = y 2 P(y) – = 0 + 0.24 + 1.76 + 2.16 + 0.64 - 2 2 2 = 4.8 – 4 = 0.8

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y01234 y - -2012 (y- ) 2 41014 P(y)0.040.240.440.240.04 (y- ) 2 P(y) 0.160.240 0.16 E(y)= y P(y) =0+0.24+0.88+0.72+0.16 = 2.00 2 = (y – P(y) = 0.16 + 0.24 + 0 + 0.24 + 0.16 2 = 0.8 Alternatively using the other formula for Variance

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