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A random variable $\mathrm{X}$ has the following distribution.
\begin{array}{|c|c|c|c|c|c|c|}
\hline \boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}} & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline \boldsymbol{P}\left(\boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}}\right) & 0.1 & \mathrm{k} & 0.2 & 2 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}
Then the variance of this distribution is
Options:
\begin{array}{|c|c|c|c|c|c|c|}
\hline \boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}} & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline \boldsymbol{P}\left(\boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}}\right) & 0.1 & \mathrm{k} & 0.2 & 2 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}
Then the variance of this distribution is
Solution:
1452 Upvotes
Verified Answer
The correct answer is:
2.16
\begin{array}{|c|c|c|c|c|}
\hline \boldsymbol{x} & \mathbf{P}(\boldsymbol{x}) & \boldsymbol{x}^2 & \boldsymbol{x}^2 \cdot \mathbf{P}(\boldsymbol{x}) & \boldsymbol{x} \cdot \mathbf{P}(\boldsymbol{x}) \\
\hline-2 & 0.1 & 4 & 0.4 & -0.2 \\
-1 & \mathrm{~K} & 1 & \mathrm{~K} & -\mathrm{K} \\
0 & 0.2 & 0 & 0 & 0 \\
1 & 2 \mathrm{~K} & 1 & 2 \mathrm{~K} & 2 \mathrm{~K} \\
2 & 3 \mathrm{~K} & 4 & 12 \mathrm{~K} & 6 \mathrm{~K} \\
3 & \mathrm{~K} & 9 & 9 \mathrm{~K} & 3 \mathrm{~K} \\
\hline
\end{array}
$\operatorname{Var}(x)=\sum\left[x^2 \cdot P(x)\right]-\left(\mu_x\right)^2$
Now $\left(\mu_x\right)^2=\left[\sum(x \cdot \mathrm{P}(x)]^2=(10 \mathrm{~K}-0.2)^2\right.$
also $\Sigma \mathrm{P}=1 \Rightarrow 7 \mathrm{~K}+0.3=1 \Rightarrow \mathrm{K}=0.1$
$\left(\mu_x\right)^2=0.64$
$\operatorname{Var}(x)=(24 \mathrm{~K}+0.4)-0.64=2.16$
\hline \boldsymbol{x} & \mathbf{P}(\boldsymbol{x}) & \boldsymbol{x}^2 & \boldsymbol{x}^2 \cdot \mathbf{P}(\boldsymbol{x}) & \boldsymbol{x} \cdot \mathbf{P}(\boldsymbol{x}) \\
\hline-2 & 0.1 & 4 & 0.4 & -0.2 \\
-1 & \mathrm{~K} & 1 & \mathrm{~K} & -\mathrm{K} \\
0 & 0.2 & 0 & 0 & 0 \\
1 & 2 \mathrm{~K} & 1 & 2 \mathrm{~K} & 2 \mathrm{~K} \\
2 & 3 \mathrm{~K} & 4 & 12 \mathrm{~K} & 6 \mathrm{~K} \\
3 & \mathrm{~K} & 9 & 9 \mathrm{~K} & 3 \mathrm{~K} \\
\hline
\end{array}
$\operatorname{Var}(x)=\sum\left[x^2 \cdot P(x)\right]-\left(\mu_x\right)^2$
Now $\left(\mu_x\right)^2=\left[\sum(x \cdot \mathrm{P}(x)]^2=(10 \mathrm{~K}-0.2)^2\right.$
also $\Sigma \mathrm{P}=1 \Rightarrow 7 \mathrm{~K}+0.3=1 \Rightarrow \mathrm{K}=0.1$
$\left(\mu_x\right)^2=0.64$
$\operatorname{Var}(x)=(24 \mathrm{~K}+0.4)-0.64=2.16$
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