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X Has the Following Probability Distribution P(X) X1234P(X).1.5.2.2\begin{array} { l l l l l } \mathrm { X } & 1 & 2 & 3 & 4 \\\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2\end{array}

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X has the following probability distribution P(X) : X1234P(X) .1.5.2.2\begin{array} { l l l l l } \mathrm { X } & 1 & 2 & 3 & 4 \\\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2\end{array}
Compute the expected value of X.

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