Variances add, even for a difference

The mean $\mu_X = \sum x \cdot P(x)$ locates the distribution. The variance measures how far the values sit from that center, weighting each squared distance by its probability:

$$\sigma_X^2 = \text{Var}(X) = \sum (x - \mu_X)^2 \cdot P(x), \qquad \sigma_X = \sqrt{\sigma_X^2}.$$

For the pets distribution ($P(0) = 0.30$, $P(1) = 0.35$, $P(2) = 0.20$, $P(3) = 0.10$, $P(4) = 0.05$, with $\mu_X = 1.25$):

1. $(0 - 1.25)^2(0.30) = 0.46875$
2. $(1 - 1.25)^2(0.35) = 0.021875$
3. $(2 - 1.25)^2(0.20) = 0.1125$
4. $(3 - 1.25)^2(0.10) = 0.30625$
5. $(4 - 1.25)^2(0.05) = 0.378125$

The variance is the total, $1.2875$ squared pets, and the standard deviation is $\sqrt{1.2875} \approx 1.13$ pets. Only the standard deviation carries the variable's own units, which is why interpretations are written with it: household pet counts typically fall about 1.13 pets from the mean of 1.25.

For constants $a$ and $b$, the linear transformation $Y = aX + b$ has

$$\mu_Y = a\mu_X + b, \qquad \sigma_Y = |a| \sigma_X, \qquad \sigma_Y^2 = a^2 \sigma_X^2.$$

Adding a constant slides every value by the same amount, so the distances between values, and therefore the spread, are untouched. Multiplying stretches those distances by $|a|$. The absolute value matters: multiplying by $-2$ flips the distribution and doubles its spread, and a standard deviation is never negative.

If every worker in a plant receives a flat 50 dollar bonus, the mean pay rises by 50 and the standard deviation is unchanged. If instead every worker's pay is multiplied by 1.05, both the mean and the standard deviation rise by 5%. Confusing the two, most often by adding a constant and then adjusting the spread, is the standard error here.

For any two random variables, means combine the obvious way:

$$E(X \pm Y) = E(X) \pm E(Y).$$

Spread is the part that surprises. When $X$ and $Y$ are independent,

$$\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) \quad \text{and} \quad \text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y).$$

Variances add in both cases. Subtracting two independent quantities produces something more variable than either, because each contributes its own uncertainty to the result. If variances subtracted, a difference of two variables with equal variances would have no variability at all, which is plainly false: the difference between two independent test scores is not a constant.

Suppose $X$, a machine's fill weight, has mean 500 g and standard deviation 100 g, and $Y$, a competitor's, has mean 480 g and standard deviation 60 g, independently. Then $E(X - Y) = 20$ g and

$$\sigma_{X-Y} = \sqrt{100^2 + 60^2} = \sqrt{13600} \approx 116.6 \text{ g}.$$

Two failed routes: subtracting variances gives $\sqrt{10000 - 3600} = 80$, and adding standard deviations gives $160$. Standard deviations never add; only variances do, and only under independence.

$2X$ and $X_1 + X_2$ have the same mean and different spreads, and the difference is the whole point:

1. $2X$ takes one observation and doubles it, so $\sigma_{2X} = 2\sigma_X$. With $\sigma_X = 100$, that is 200.
2. $X_1 + X_2$ takes two independent observations and adds them, so $\sigma^2 = \sigma_X^2 + \sigma_X^2 = 2\sigma_X^2$ and $\sigma = \sqrt{2}\,\sigma_X \approx 141.4$.

Two independent draws partly cancel each other's departures from the mean, so their sum is less variable than one draw counted twice. In context: one customer's order doubled swings twice as far as that customer's order; two customers' orders together swing only about 1.41 times as far.

Independence is the condition on all of this. Adding variances for dependent variables is wrong, and the AP course does not supply a rule for that case, so an independence claim has to be checked or given before any variance is combined.