Math Core

A probability distribution tells you everything about a random variable, but you often want a quick summary: what value do you expect on average, and how much do results typically vary? The mean and standard deviation of a random variable answer those questions, and they drive decisions about games, insurance and business plans.

The mean (expected value)

The mean of a random variable is a weighted average of its values, where each value is weighted by its probability.

Mean of a discrete random variable

$\mu_X = E(X) = \sum x_i \, p_i = x_1 p_1 + x_2 p_2 + \dots + x_k p_k.$

The mean is also called the expected value. It is the long-run average of $X$ over many, many repetitions of the random process.

The expected value does not have to be a possible value of $X$ . A family can't own $1.8$ cars, but $1.8$ can still be the long-run average number of cars per household.

The standard deviation

The variance of $X$ is the weighted average of squared deviations from the mean, and the standard deviation is its square root:

$\sigma_X^2 = \sum (x_i - \mu_X)^2 \, p_i, \qquad \sigma_X = \sqrt{\sigma_X^2}.$

The standard deviation measures how much the values of $X$ typically vary from the mean, in the same units as $X$ .

Worked example: Cars per household

Let $X$ = the number of cars owned by a randomly selected household in a town.

$x$	0	1	2	3	4
$P(X = x)$	0.08	0.32	0.38	0.16	0.06

Find the mean and standard deviation of $X$ , and interpret the mean.

Solution. Mean:

$\mu_X = 0(0.08) + 1(0.32) + 2(0.38) + 3(0.16) + 4(0.06) = 0 + 0.32 + 0.76 + 0.48 + 0.24 = 1.80.$

Variance, using deviations from $1.8$ :

\begin{aligned} \sigma_X^2 &= (0 - 1.8)^2(0.08) + (1 - 1.8)^2(0.32) + (2 - 1.8)^2(0.38) + (3 - 1.8)^2(0.16) + (4 - 1.8)^2(0.06) \\ &= 0.2592 + 0.2048 + 0.0152 + 0.2304 + 0.2904 \\ &= 1.00. \end{aligned}

So $\sigma_X = \sqrt{1.00} = 1.00$ car.

Interpretation. If you selected many, many households at random from this town, the average number of cars per household would be about $1.8$ . The number of cars for a randomly chosen household typically varies from the mean of $1.8$ by about $1$ car.

Tip

Your calculator can do this. On a TI-84, put the values in L1 and the probabilities in L2, then run 1-Var Stats with List: L1 and FreqList: L2. The output $\bar{x}$ is $\mu_X$ and $\sigma x$ is $\sigma_X$ . On the AP exam, still show the formula with at least the first few terms substituted.

Expected value and decisions

Expected value tells you whether a game or a policy pays off in the long run.

Worked example: Is the game worth playing?

A carnival game costs $5 to play. You win $20 with probability $0.1$ , win $5 with probability $0.3$ , and win nothing otherwise. Let $X$ = your net gain on one play. Find $E(X)$ .

Solution. Subtract the $5 cost from each prize to get net gains.

Net gain $x$ (dollars)	15	0	$-5$
$P(X = x)$	0.1	0.3	0.6

$E(X) = 15(0.1) + 0(0.3) + (-5)(0.6) = 1.5 + 0 - 3 = -1.5.$

On average you lose $1.50 per play over many plays. The game operator makes about $1.50 per play in the long run.

Worked example: Insurance

An insurance company charges $300 per year for a policy that pays $10,000 if a certain event occurs. The event happens to $2\%$ of policyholders each year. What is the company's expected profit per policy?

Solution. Profit is $300$ if no claim and $300 - 10{,}000 = -9{,}700$ if there is a claim.

$E(\text{profit}) = 300(0.98) + (-9{,}700)(0.02) = 294 - 194 = 100.$

The company expects to earn about $100 per policy on average. It will lose big on some policies, but across thousands of policies, the law of large numbers makes the average profit close to $100.

Common mistake

Don't interpret the expected value as what will happen on one trial, or as the "most likely" value. It is a long-run average. Say "over many repetitions, the average … would be about …" and include context. Also remember to square the deviations and multiply by probabilities, not divide by $n$ or $n - 1$ as for sample data.