Math Core

Real quantities are often built from other random quantities: a total time is the sum of several steps, a profit is revenue minus cost, and a fare is a fixed fee plus a charge per mile. This lesson shows how the mean and standard deviation change when you transform a random variable or combine two of them.

Linear transformations

Suppose $Y = a + bX$ . Adding a constant $a$ shifts every value, so it shifts the mean but doesn't change the spread. Multiplying by $b$ stretches every value and every distance between values, so it multiplies both the mean and the standard deviation.

Linear transformation of a random variable

If $Y = a + bX$ , then

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

The shape of the distribution stays the same (for $b > 0$ ).

Worked example: Taxi fare

A taxi company charges a $3 pickup fee plus $2.50 per mile. The length $X$ of a randomly chosen trip has mean $6$ miles and standard deviation $2.4$ miles. Find the mean and standard deviation of the fare $F = 3 + 2.5X$ .

Solution.

$\mu_F = 3 + 2.5(6) = 18, \qquad \sigma_F = 2.5(2.4) = 6.$

The mean fare is $18, with a standard deviation of $6. The $3 fee raises every fare by the same amount, so it doesn't affect the spread.

Sums and differences

Now combine two random variables $X$ and $Y$ .

Means always add or subtract. For any random variables,

$\mu_{X+Y} = \mu_X + \mu_Y, \qquad \mu_{X-Y} = \mu_X - \mu_Y.$

Variances add, for both sums and differences, when $X$ and $Y$ are independent. This surprises many students. Combining two sources of randomness always makes the result more variable, whether you add or subtract them.

Variances add for independent random variables

If $X$ and $Y$ are independent,

$\sigma_{X+Y}^2 = \sigma_X^2 + \sigma_Y^2, \qquad \sigma_{X-Y}^2 = \sigma_X^2 + \sigma_Y^2.$

So $\sigma_{X \pm Y} = \sqrt{\sigma_X^2 + \sigma_Y^2}$ . Standard deviations do not add; only variances do.

Why does the difference have the larger spread? Think of the extremes. $X - Y$ is largest when $X$ is big and $Y$ is small, and smallest when $X$ is small and $Y$ is big, so its range is the range of $X$ plus the range of $Y$ .

Worked example: Getting ready in the morning

The time $X$ a student spends showering has mean $8$ minutes and standard deviation $2$ minutes. The time $Y$ spent getting dressed has mean $6$ minutes and standard deviation $1.5$ minutes. Assume $X$ and $Y$ are independent. Find the mean and standard deviation of the total time $T = X + Y$ .

Solution.

$\mu_T = 8 + 6 = 14 \text{ minutes}.$

$\sigma_T^2 = 2^2 + 1.5^2 = 4 + 2.25 = 6.25, \qquad \sigma_T = \sqrt{6.25} = 2.5 \text{ minutes}.$

Notice $\sigma_T = 2.5$ , not $2 + 1.5 = 3.5$ .

Sums of several independent copies

If $X_1, X_2, \dots, X_n$ are independent and each has mean $\mu$ and standard deviation $\sigma$ , the total has mean $n\mu$ and standard deviation $\sqrt{n}\,\sigma$ . This is different from $nX$ (one value multiplied by $n$ ), which has standard deviation $n\sigma$ . Several independent values partly cancel each other's randomness; one value multiplied does not.

Combining normal random variables

If $X$ and $Y$ are independent and both normally distributed, then $X + Y$ and $X - Y$ are also normally distributed. That lets you compute probabilities with $z$ -scores.

Worked example: Which takes longer?

Using the morning routine above, suppose both times are approximately normal. Find the probability that the student spends less time showering than getting dressed on a randomly chosen morning.

Solution. Let $D = X - Y$ . We want $P(D \lt 0)$ .

$\mu_D = 8 - 6 = 2, \qquad \sigma_D = \sqrt{2^2 + 1.5^2} = 2.5.$

$D$ is normal, so

$z = \dfrac{0 - 2}{2.5} = -0.8, \qquad P(D \lt 0) = P(Z \lt -0.8) \approx 0.2119.$

On about $21\%$ of mornings, dressing takes longer than showering.

Common mistake

The most common mistake is subtracting variances (or standard deviations) for a difference. Variances always add for independent random variables. A second mistake is adding standard deviations directly; add the variances, then take the square root. And the variance rule requires independence: state that assumption, and question it when it's doubtful.

Practice

Practice 1

A random variable $X$ has mean $10$ and standard deviation $2$ . Let $Y = 4X - 3$ . Find the mean of $Y$ .