The variance of a probability distribution is:
[tex]\sigma^2=\Sigma x^2p-\mu^2[/tex]
where p is the probability and mu is the mean of the ditribution.
Before we begin we need to find the mean of the distribution, using:
[tex]\mu=\Sigma xp[/tex]
Then the mean is:
[tex]\begin{gathered} \mu=(-50)(0.029)+(-30)(0.121)+(-10)(0.210)+(10)(0.485)+(30)(0.145)+(50)(0.010) \\ =2.52 \end{gathered}[/tex]
hence the mean is 2.52.
Now we can calculate the variance:
[tex]\begin{gathered} \sigma^2=(-50)^2(0.029)+(-30)^2(0.121)+(-10)^2(0.210)+(10)^2(0.485)+(30)^2(0.145)+(50)^2(0.010)-2.52^2 \\ =400.04 \end{gathered}[/tex]
Therefore, the answer is C.
