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Question 19

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 Using the probability distribution listed, the mean would be 1.6X0123P(X)0.20.10.30.4\begin{array}{l}\text { Using the probability distribution listed, the mean would be } 1.6 \text {. }\\\begin{array} { l | l l l l } \boldsymbol { X } & 0 & 1 & 2 & 3 \\\hline \boldsymbol { P } ( \boldsymbol { X } ) & 0.2 & 0.1 & 0.3 & 0.4\end{array}\end{array}

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