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The Probability Distribution for X Is as Follows x1012p(x)0.10.250.550.1\begin{array} { l | r r r r } x & - 1 & 0 & 1 & 2 \\\hline p ( x ) & 0.1 & 0.25 & 0.55 & 0.1\end{array}

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The probability distribution for X is as follows: x1012p(x)0.10.250.550.1\begin{array} { l | r r r r } x & - 1 & 0 & 1 & 2 \\\hline p ( x ) & 0.1 & 0.25 & 0.55 & 0.1\end{array} Find the expected value of Y = X + 10.

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