X has the probability distribution: $\begin{array} { l r r r r r }
\mathrm { X } & - 2 & - 1 & 9 & 1 & 2 \
\mathrm { P } ( \mathrm { X } ) & .2 & .2 & .2 & .2 & .2
\end{array}$
Compute the expected value of X.
A)1.3
B)1.0
C)2.4
D)1.8

Question

X has the probability distribution: $\begin{array} { l r r r r r } 
\mathrm { X } & - 2 & - 1 & 9 & 1 & 2 \
\mathrm { P } ( \mathrm { X } ) & .2 & .2 & .2 & .2 & .2
\end{array}$
Compute the expected value of X.
A)1.3
B)1.0
C)2.4
D)1.8

Quizplus · Accepted Answer

The expected value of X, denoted as E(X), is calculated by taking the sum of each possible outcome multiplied by its corresponding probability:
E(X) = (-2)(0.2) + (-1)(0.2) + (9)(0.2) + (1)(0.2) + (2)(0.2) = 1.8
Therefore, the correct choice is D.

X Has the Probability Distribution X−2−1912P(X).2.2.2.2.2\begin{array} { l r r r r r } \mathrm { X } & - 2 & - 1 & 9 & 1 & 2 \\\mathrm { P } ( \mathrm { X } ) & .2 & .2 & .2 & .2 & .2\end{array}XP(X)​−2.2​−1.2​9.2​1.2​2.2​

X Has the Probability Distribution $\begin{array} { l r r r r r } \mathrm { X } & - 2 & - 1 & 9 & 1 & 2 \\\mathrm { P } ( \mathrm { X } ) & .2 & .2 & .2 & .2 & .2\end{array}$