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 variance value of X.
A)1.58
B).955
C).912
D).625

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 variance value of X.
A)1.58
B).955
C).912
D).625

Quizplus · Accepted Answer

To compute the variance of X, we first need to find the expected value or mean of X, which we can compute as follows:

E(X) = 1(.1) + 2(.5) + 3(.2) + 4(.2) = 2.3

Next, we can use the formula for variance:

Var(X) = E(X^2) - [E(X)]^2

To find E(X^2), we can use the formula:

E(X^2) = 1^2(.1) + 2^2(.5) + 3^2(.2) + 4^2(.2) = 8.1

Therefore,

Var(X) = 8.1 - 2.3^2 = 0.912

So, the answer is (C).

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