Give the mean of the following probability distribution? $$\begin{array} { | l | l | }
\hline x & P ( x ) \
\hline 1 & 0.20 \
\hline 2 & 0.10 \
\hline 3 & 0.35 \
\hline 4 & 0.05 \
\hline 5 & 0.30 \
\hline
\end{array}$$
A) 1.20
B) 1.46
C) 3.15
D) 2.13

Question

Give the mean of the following probability distribution? $$\begin{array} { | l | l | } 
\hline x & P ( x ) \
\hline 1 & 0.20 \
\hline 2 & 0.10 \
\hline 3 & 0.35 \
\hline 4 & 0.05 \
\hline 5 & 0.30 \
\hline
\end{array}$$
A) 1.20
B) 1.46
C) 3.15
D) 2.13

Quizplus · Accepted Answer

To find the mean of a probability distribution, we multiply each outcome by its probability and sum the products. In this case, we have: \begin{align*}
	ext{Mean} &= (1)(0.20) + (2)(0.10) + (3)(0.35) + (4)(0.05) + (5)(0.30) \
&= 0.20 + 0.20 + 1.05 + 0.20 + 1.50 \
&= 3.15
\end{align*}
Therefore, the best choice is C.

Give the Mean of the Following Probability Distribution A) 120 B) 1

Give the Mean of the Following Probability Distribution A) 120
B) 1