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

Question

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

Quizplus · Accepted Answer

The mean of a probability distribution is the expected value, which is the sum of each possible outcome multiplied by its probability. In this case, the mean is (0 * 0.2) + (1 * 0.1) + (2 * 0.3) + (3 * 0.4) = 1.6.