Combining Normal random variables: totals, multiples and the key result

This section starts the A Level unit Statistics 3 (WST03/01), which counts towards Further Mathematics.

On every recent S3 paper, the longest question, worth 15 or 16 of the 75 marks, has been about combining Normal random variables: a total, a difference, then a mixture of the two.

This section builds the one result behind all of it and uses it for totals and multiples; the next section deals with differences and with the traps in the closing part of that question.

You need the S1 Normal work (standardising and the tables) and the rules $\mathrm{E}(aX + b) = a\mathrm{E}(X) + b$ and $\mathrm{Var}(aX + b) = a^{2}\mathrm{Var}(X)$.

A shop sells apples whose masses vary: some 150 g, some 170 g, most close to 160 g.

Put five apples in a bag.

The bag's total mass also varies, and the questions a packer asks are about the total: how likely is a bag to be over 830 g?

To answer, you need the distribution of a sum of random variables, not of one.

Two things about a total are fairly obvious.

Its mean should be five times the mean of one apple.

And it should vary more than one apple does, because five apples' worth of variation is piling up.

What is less obvious is how much more.

Heavy and light apples partly cancel in a bag, so the spread does not grow five times.

The rule that captures this exactly is that variances add, while standard deviations do not.

The other surprise is shape.

If each apple's mass is Normal, then the total is Normal too.

That is a special property of the Normal distribution, and it is what lets every S3 question finish with the Normal tables.

Learn it in this form.

If $X \sim \mathrm{N}(\mu_x, \sigma_x^{2})$ and $Y \sim \mathrm{N}(\mu_y, \sigma_y^{2})$ independently, then

Three things to read in it:

1. Means behave like ordinary numbers. Multiply, add and subtract them exactly as the expression says.

2. Variances are always added, even for $aX - bY$, and each is multiplied by the square of its coefficient. A variance can never be negative, and subtracting a random amount adds uncertainty; it never removes it.

3. Independence is required for the variance rule. The mean rule holds whatever the dependence.

Learn it. This result is a formula you are expected to know: it is not printed in the Mathematical Formulae and Statistical Tables booklet in this form.

The booklet's Statistics S3 section prints only the variance part, $\mathrm{Var}(aX \pm bY) = a^{2}\mathrm{Var}(X) + b^{2}\mathrm{Var}(Y)$ for independent $X$ and $Y$.

No proofs are required.

The result extends to any number of independent Normal variables: $X_1 + X_2 + \dots + X_n$ is Normal, with mean the sum of the means and variance the sum of the variances.

The mean rule comes straight from the definition of expectation: $\mathrm{E}(aX + bY) = a\mathrm{E}(X) + b\mathrm{E}(Y)$, because an expectation is a weighted sum and sums can be split.

For the variance, write $X' = X - \mu_x$ and $Y' = Y - \mu_y$, the deviations from the means.

Then

The first and last terms are $a^{2}\sigma_x^{2}$ and $b^{2}\sigma_y^{2}$.

The middle term measures whether $X$ and $Y$ tend to be above their means together.

For independent variables $\mathrm{E}(X'Y') = \mathrm{E}(X')\mathrm{E}(Y') = 0 \times 0 = 0$, so it vanishes.

With $-b$ in place of $b$ the middle term changes sign, but it is still zero, and $(-b)^{2} = b^{2}$: that is why the variances add for a difference too.

A small exploration with dice shows the rule at work.

One fair die has variance $\frac{35}{12} \approx 2.92$.

The total of two separate dice can be listed in full (36 equally likely outcomes) and has variance $\frac{35}{6} \approx 5.83$, exactly twice as much.

Doubling the score on one die gives 2, 4, ..., 12 with variance $4 \times \frac{35}{12} \approx 11.67$, four times as much.

Same mean, 7, in both cases, very different spread.

That the shape stays Normal for Normal variables is a deeper fact, proved at university with moment generating functions.

This is the idea that decides more marks in this unit than any other.

Suppose square tiles have side $S \sim \mathrm{N}(20, 0.3^{2})$ cm.

• Four separate tiles laid in a row. The row's length is $S_1 + S_2 + S_3 + S_4$, a sum of four independent sides. Mean $4 \times 20 = 80$; variance $0.09 + 0.09 + 0.09 + 0.09 = 4 \times 0.09 = 0.36$.

• The perimeter of one tile. That is $4S$: one random side, used four times. Mean $4 \times 20 = 80$; variance $4^{2} \times 0.09 = 16 \times 0.09 = 1.44$.

The means agree.

The variances do not: $n\sigma^{2}$ for $n$ separate copies against $n^{2}\sigma^{2}$ for one copy multiplied by $n$.

The sketch below shows the two distributions.

The reason is the cancelling described at the start.

In a row of four tiles a large tile is often next to a small one, so errors partly cancel.

In one tile's perimeter there is nothing to cancel: if the side is too big, all four sides are too big.

So before writing any variance, ask: is each item a fresh, independent copy, or is the same item counted several times? The words of the question decide.

"The total of 5 apples" means five copies.

"Three times the mass of an apple" means $3X$ for one apple.

1. Define the new variable in words and symbols, for example $T = A_1 + A_2 + \dots + A_5$.

2. Mean: combine the means exactly as the expression says.

3. Variance: add the variances, each times its coefficient squared; count separate copies one by one.

4. Write $T \sim \mathrm{N}(\text{mean}, \text{variance})$. This line usually earns M1 A1.

5. Standardise with the standard deviation, the square root of the variance: $Z = \frac{T - \text{mean}}{\text{sd}}$.

6. Use the tables: $\Phi(z)$ to 4 d.p., at $z$ rounded to 2 d.p. A calculator's Normal function gives a slightly different value; both are accepted when the method is shown, which is why a mark scheme gives a range.

A probability about a total is a statement about bags, rows, loads.

"$\mathrm{P}(T > 830) = 0.131$" says about 13% of bags of five apples will weigh more than 830 g, so a packer who labels bags "800 g" gives away more than 30 g of extra fruit in roughly one bag in eight (and some extra fruit in about half of all bags, since the mean is exactly 800 g).

Notice how the spread of a total grows only like $\sqrt{n}$ while its mean grows like $n$: relative to its size, a large total is far more predictable than one item.

This is the idea behind the next S3 topic, the distribution of the sample mean $\bar{X}$, where $\mathrm{Var}(\bar{X}) = \frac{\sigma^{2}}{n}$ comes from exactly this rule.

The same algebra appears across science and finance.

In manufacturing, tolerances stack: the gap left when parts are assembled is a linear combination of the parts' sizes.

In finance, the risk of a portfolio is a combination of the variances of its holdings (with a covariance term when they are not independent, the term that vanished above).

In S3 itself, the sample mean, the two-sample test and the standard error all rest on "variances add".

1. The random variables $X \sim \mathrm{N}(20, 9^{2})$ and $Y \sim \mathrm{N}(35, 4^{2})$ are independent. Write down the distribution of $3X + 2Y$, giving its mean and variance. (2)

Total: [2]

2. A student says: "$X_1 + X_2$ and $2X$ have the same distribution, because both have mean $2\mu$." Explain why the student is wrong, stating the variance of each. (2)

Total: [2]

3. The mass of an egg, $E$ grams, is modelled as $E \sim \mathrm{N}(62, 4^{2})$. Six eggs, chosen at random, are packed in a box whose mass is $B \sim \mathrm{N}(30, 2^{2})$ grams, independent of the eggs.

(a) Find the distribution of the total mass of a full box. (2)

(b) Using standardisation, find the probability that a full box has a total mass between 390 g and 420 g. (4)

Total: [6]

4. The volume of coffee poured by a machine into a cup, $C$ ml, is modelled as $C \sim \mathrm{N}(240, 6^{2})$. The machine fills 25 cups, chosen at random, from one container.

Using standardisation, show that the probability that the 25 cups take more than 6045 ml in total is 0.0668 (4)

Total: [4]

5. Slices of bread have mass $S \sim \mathrm{N}(35, 2^{2})$ grams and portions of filling have mass $F \sim \mathrm{N}(60, 5^{2})$ grams. A sandwich is made from two slices of bread and one portion of filling, all chosen at random and independent.

(a) Find the probability that a sandwich has mass less than 120 g. (4)

A pack holds three sandwiches.

(b) Find the probability that the total mass of the sandwiches in a pack is more than 405 g. (4)

(c) Explain why the variance in part (a) uses $2 \times 2^{2}$ for the bread and not $2^{2} \times 2^{2}$. (1)

Total: [9]

6. The random variable $X \sim \mathrm{N}(\mu, 5^{2})$. The random variable $T$ is the sum of four independent observations of $X$. Given that $\mathrm{P}(T > 100) = 0.0228$, find the value of $\mu$. (3)

Total: [3]

7. Explain what is wrong with the claim "if $X$ and $Y$ are independent then $\mathrm{Var}(X - Y) = \mathrm{Var}(X) - \mathrm{Var}(Y)$". (2)

Total: [2]