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

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

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2.5 EV = (1)(.1) + (...

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