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The Probability Distribution for X ,Daily Demand of a Particular x1234p(x)0.050.420.440.09\begin{array} { l | r r r r } x & 1 & 2 & 3 & 4 \\\hline p ( x ) & 0.05 & 0.42 & 0.44 & 0.09\end{array}

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The probability distribution for X ,daily demand of a particular newspaper at a local newsagency,( in hundreds) is as follows: x1234p(x)0.050.420.440.09\begin{array} { l | r r r r } x & 1 & 2 & 3 & 4 \\\hline p ( x ) & 0.05 & 0.42 & 0.44 & 0.09\end{array} a. Find and interpret the expected value of X.
b. Find V(X).
c. Find σ\sigma .

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a. E[X] = 2.57 in hundreds of newspapers...

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