X has the following probability distribution P(X): $\begin{array} { l l l l l }
\mathrm { X } & 1 & 2 & 3 & 4 \
\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2
\end{array}$
Compute the expected value of X.
A)2.5
B)1.0
C)1.6
D)0.6

Question

X has the following probability distribution P(X): $\begin{array} { l l l l l } 
\mathrm { X } & 1 & 2 & 3 & 4 \
\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2
\end{array}$
Compute the expected value of X.
A)2.5
B)1.0
C)1.6
D)0.6

Quizplus · Accepted Answer

The expected value E(X) is calculated as: $E(X) = 1(0.1) + 2(0.5) + 3(0.2) + 4(0.2) = 0.1 + 1.0 + 0.6 + 0.8 = 2.5$.

X Has the Following Probability Distribution P(X) X1234P(X).1.5.2.2\begin{array} { l l l l l } \mathrm { X } & 1 & 2 & 3 & 4 \\\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2\end{array}XP(X)​1.1​2.5​3.2​4.2​

X Has the Following Probability Distribution P(X) $\begin{array} { l l l l l } \mathrm { X } & 1 & 2 & 3 & 4 \\\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2\end{array}$