1 Answer

Jmuc · Answer 1 · 2023-12-02T15:24:27+0000

The variance of a probability distribution is:

$\sigma^2=\Sigma x^2p-\mu^2$

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:

$\mu=\Sigma xp$

Then the mean is:

$\begin{gathered} \mu=(-50)(0.029)+(-30)(0.121)+(-10)(0.210)+(10)(0.485)+(30)(0.145)+(50)(0.010) \\ =2.52 \end{gathered}$

hence the mean is 2.52.

Now we can calculate the variance:

$\begin{gathered} \sigma^2=(-50)^2(0.029)+(-30)^2(0.121)+(-10)^2(0.210)+(10)^2(0.485)+(30)^2(0.145)+(50)^2(0.010)-2.52^2 \\ =400.04 \end{gathered}$

Therefore, the answer is C.

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