A random variable $X$ has the following probability distribution
The variance of this random variable is

Question

A random variable $X$ has the following probability distribution The variance of this random variable is

Aicharya · Accepted Answer

We have,

We know that sum of probability $=1$. $ \begin{aligned} & \Sigma p\left(X=x_i\right)=1 \\ & \begin{array}{l} \Rightarrow \quad \frac{1}{6}+k+\frac{1}{4}+k+\frac{1}{6}=1 \\ \Rightarrow \quad 2 k=1-\frac{7}{12} \Rightarrow k=\frac{5}{24} \end{array} \\ & \text { Now, } E(X)=\Sigma x_i p_i \\ & \quad=-2\left(\frac{1}{6}\right)-1\left(\frac{5}{24}\right)+0\left(\frac{1}{4}\right)+1\left(\frac{5}{24}\right)+2\left(\frac{1}{6}\right) \\ & \begin{array}{c} E(x)=0 \\ \text { Now, } E\left(X^2\right)=\Sigma x_i^2 p i \\ \quad=(-2)^2\left(\frac{1}{6}\right)+(-1)^2\left(\frac{5}{24}\right)+0\left(\frac{1}{4}\right) \\ \quad=\frac{4}{6}+\frac{5}{24}+\frac{5}{24}+\frac{4}{6}=\frac{16+10+16}{24}=\frac{42}{24}=\frac{7}{4} \end{array} \\ & \begin{array}{l} \operatorname{var}(x)=E(x)^2-[E(x)]^2 \\ \operatorname{var}(x)=\frac{7}{4}-0=\frac{7}{4} \end{array} \end{aligned} $

A random variable $X$ has the following probability distribution The variance of this random variable is

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