The number of golf balls ordered by customers of a pro shop has the following probability distribution. $\begin{array}{|c|c|}
\hline \mathrm{x} & \mathrm{P}(\mathrm{x}) \
\hline 3 & 0.14 \
6 & 0.29 \
9 & 0.36 \
12 & 0.11 \
15 & 0.10 \
\hline
\end{array}$

A) $\mu = 8.22$
B) $\mu = 9$
C) $\mu = 9.3$
D) $\mu = 5.55$

Question

The number of golf balls ordered by customers of a pro shop has the following probability distribution. $\begin{array}{|c|c|}
\hline \mathrm{x} & \mathrm{P}(\mathrm{x}) \
\hline 3 & 0.14 \
6 & 0.29 \
9 & 0.36 \
12 & 0.11 \
15 & 0.10 \
\hline
\end{array}$

A) $\mu = 8.22$
B) $\mu = 9$
C) $\mu = 9.3$
D) $\mu = 5.55$

Quizplus · Accepted Answer

The expected value (mean) $\mu$ is calculated as $\mu = \sum x \cdot P(x) = 3(0.14) + 6(0.29) + 9(0.36) + 12(0.11) + 15(0.10) = 8.22$.

The Number of Golf Balls Ordered by Customers of a Pro