$\begin{array}{|c|c|}
\hline \mathrm{x} & \mathrm{P}(\mathrm{x}) \
\hline 0 & 0.26 \
1 & 0.11 \
2 & 0.16 \
3 & 0.05 \
4 & 0.42 \
\hline
\end{array}$

A) $\mu = 2.16$
B) $\mu = 2.26$
C) $\mu = 2.42$
D) $\mu = 2.52$

Question

$\begin{array}{|c|c|}
\hline \mathrm{x} & \mathrm{P}(\mathrm{x}) \
\hline 0 & 0.26 \
1 & 0.11 \
2 & 0.16 \
3 & 0.05 \
4 & 0.42 \
\hline
\end{array}$

A) $\mu = 2.16$
B) $\mu = 2.26$
C) $\mu = 2.42$
D) $\mu = 2.52$

Quizplus · Accepted Answer

The expected value $\mu$ is calculated as $\mu = \sum x \cdot P(x) = 0 \cdot 0.26 + 1 \cdot 0.11 + 2 \cdot 0.16 + 3 \cdot 0.05 + 4 \cdot 0.42 = 2.26$.

A) μ=2.16\mu = 2.16μ=2.16 B) μ=2.26\mu = 2.26μ=2.26 C) μ=2.42\mu = 2.42μ=2.42

A) $\mu = 2.16$ B) $\mu = 2.26$ C) $\mu = 2.42$