A **random variable** is a variable that whose value can change under different outcomes. For example, the result of flipping a standard 6-sided die is a random variable that takes each of the values from 1 to 6 with probability \(\frac{1}{6}.\)

A random variable contains a lot of information. In Expected Value, we learned that the weighted average of all possible outcomes gives one way of understanding the random variable. Recall that \( E[X] = \mu\).

The **variance** of a random variable measures how far the random variable deviates from its mean, by calculating the expected value of the square deviation from the mean. In mathematical symbols, we have

\[ Var (X) = E[ ( X - \mu)^2 ]. \]

It can also be shown, by the linearity of expectation, that

\[ Var (X) = E[X^2] - \left( E[X] \right)^2 .\]

## 1. There are \(2\) bags, containing balls numbered \(1\) through \(5\). From each bag, \(1\) ball is removed. What is the variance of the total of the two balls?

Let \(X\) be the random variable denoting the sum of these values. Then, the probability distribution of \(X\) is given by

\[\begin{array} { | l | l | l | l | l | l | l | l | l | l l | } \hline x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \\ \hline P(X=x) & \frac{1}{25} & \frac{2}{25} & \frac{3}{25} & \frac{4}{25} & \frac{5}{25} & \frac{4}{25} & \frac{3}{25} & \frac{2}{25} & \frac{1}{25} & \\ \hline \end{array} \]

We have previously calculated that \( E[X] = 6. \) As such, we can see that

\[ \begin{align} Var(X) & = E[(X - \mu)^2] \\ & = (2-6)^2 \times \frac {1}{25} + (3-6)^2 \times \frac {2}{25} + (4-6)^2 \times \frac {3}{25} \\ & + (5-6)^2 \times \frac {4}{25} + (6-6)^2 \times \frac {5}{25} + (7-6)^2 \times \frac {4}{25} \\ & + (8-6)^2 \times \frac {3}{25} + (9-6)^2 \times \frac {2}{25} + (10-6)^2 \times \frac {1}{25} \\ & = 4.

\end{align} \]

## 2. \(3\) six-sided dice are rolled. What is the variance for the number of times a \(3\) is rolled? What is the variance for the total of the dice?

Let \(Y\) be the random variable representing the number of times a \(3\) is rolled. The below table lists the probabilities of rolling different numbers of \(3\)s.

\[ \begin{array}{|c|cccc|} \hline \mbox{num 3s} & 0 & 1 & 2 & 3\\ \mbox{probability } & \frac{125}{216} & \frac{75}{216} & \frac{15}{216} & \frac{1}{216}\\ \hline \end{array} \]

The expected number of times a \(3\) is rolled is \(\frac{1}{2}\), so \(E(Y) = \frac{1}{2}\). We now wish to calculate \(E(Y^2)\).

\[ E(Y^2) = \frac{0^2 \times 125 + 1^2 \times 75 + 2^2 \times 15 + 3^2 \times 1}{216} = \frac{144}{216} = \frac{2}{3}. \]

Therefore, \(Var(Y) = \frac{2}{3} - \left(\frac{1}{2}\right)^2 = \frac{5}{12}\).

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## Comments

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TopNewestWhy do you have to square the differences?

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We are interested in measuring how spread out the data is. Squaring allows us to emphasize large differences while reducing the impact of small differences.

Variance, is defined as the second central moment, which is why we take it as \( E[ (X - \mu)^2] \). Skew, is the third central moment, in which we take cubes instead of squares, namely \( E[ ( X - \mu)^3 ] \).

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But those cubes won't always be positive....?

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