Given that
The table consists of probability distribution x and P(x) as
Explanation -
The formula for the variance is
[tex]\begin{gathered} Var=\sigma^2=\sum_{x\mathop{=}-4}^0x^2\times P(x)-\mu^2 \\ where\text{ }\mu\text{ is the standard deviation } \\ \mu=\sum_{x\mathop{=}-4}^0x\times P(x) \end{gathered}[/tex]
So on substituting the values we have
[tex]\begin{gathered} \mu=-4\times0.2+(-3)\times0.3+(-2)\times0.1+(-1)\times0.2+(0)\times0.2 \\ \mu=-0.8-0.9-0.2-0.2+0 \\ \mu=-2.1 \\ and \\ \sigma^2=16\times0.2+9\times0.3+4\times0.1+1\times0.2+0-4.41 \\ \sigma^2=6.5-4.41=2.09 \end{gathered}[/tex]
Hence, the variance is 2.09.
And the final answer is 2.09
