Linear Combinations of Normal Variables

A2 · S2 · 12 min

The expectation and variance rules tell you the mean and variance of a combination such as $X_1 + X_2 + X_3$ or $X - 2Y$ . To find a probability, you also need the shape of its distribution. For normal variables the answer is as simple as it could be: any linear combination of independent normal variables is itself normal. This turns "total mass of a lift full of people", "is one bag heavier than three small ones" and "do these parts fit together" into ordinary normal-table calculations. It is a staple long question on Paper 6.

The key facts

Key result

If $X \sim N(\mu, \sigma^2)$ then $aX + b \sim N(a\mu + b,\ a^2\sigma^2)$ .
If $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$ are independent, then

aX + bY \sim N\!\left(a\mu_1 + b\mu_2,\ a^2\sigma_1^2 + b^2\sigma_2^2\right).

In particular, if $X_1, \dots, X_n$ are independent observations of $X \sim N(\mu, \sigma^2)$ , then

X_1 + \cdots + X_n \sim N(n\mu,\ n\sigma^2).

Proofs are not required.

The mean and variance come straight from the general rules in linear combinations of random variables. The new information is the word normal: the combination keeps the bell shape. That is special. A sum of two independent binomial variables with different $p$ , for example, is not binomial, but a sum of normals is always normal.

Probabilities for a linear combination of normals

Define each random variable in words, with its distribution.
Write the event as a single combination compared with a number: "total $> 1000$ ", " $L - 3S > 0$ ", " $|X_1 - X_2| < 2$ ".
Find the mean and variance of the combination, showing each term.
State its distribution: " $T \sim N(900, 2400)$ ".
Standardise and use the table. Sketch the curve if the region is two-sided or the sign is unclear.

A single variable scaled and shifted

The mass, $X$ grams, of a chocolate bar has the distribution $N(50, 16)$ . The cost of making a bar, in cents, is $Y = 3X - 20$ . Find the probability that a randomly chosen bar costs more than $140$ cents to make.

Solution

$E(Y) = 3(50) - 20 = 130$ and $\text{Var}(Y) = 3^2(16) = 144$ , so $Y \sim N(130, 144)$ with $\sigma = 12$ .

P(Y > 140) = P\left(Z > \frac{140 - 130}{12}\right) = P(Z > 0.833) = 1 - 0.7976 = 0.202

Totals

The total of $n$ separate items has mean $n\mu$ and variance $n\sigma^2$ . Remember the distinction from the general rules: $n$ different items give $n\sigma^2$ , while $n$ times one item gives $n^2\sigma^2$ .

Total mass of a bag

The masses of apples are normally distributed with mean $150$ g and standard deviation $20$ g. A bag contains $6$ randomly chosen apples. Find the probability that the total mass of the apples in the bag exceeds $1$ kg.

Solution

Let $T = X_1 + \cdots + X_6$ , where $X_i \sim N(150, 20^2)$ independently.

$E(T) = 6 \times 150 = 900$ ; $\text{Var}(T) = 6 \times 400 = 2400$ . So $T \sim N(900, 2400)$ .

P(T > 1000) = P\left(Z > \frac{1000 - 900}{\sqrt{2400}}\right) = P(Z > 2.041) = 1 - 0.9794 = 0.0206

Using $\text{Var}(6X) = 36 \times 400$ instead would give a standard deviation of $120$ and a probability of $0.202$ : ten times too large. The bag contains six different apples, so it is a sum.

Comparisons: turn them into differences

"Find the probability that $X$ is greater than $Y$ " cannot be done by finding two separate probabilities. Instead, combine the variables into one:

P(X > Y) = P(X - Y > 0).

Then $D = X - Y$ is normal, with mean $\mu_1 - \mu_2$ and variance $\sigma_1^2 + \sigma_2^2$ (the variances add, even for a difference).

The same trick handles any comparison. " $L$ is more than three times $S$ " becomes $L - 3S > 0$ . " $X$ exceeds $Y$ by more than $5$ " becomes $X - Y > 5$ . " $A$ and $B$ differ by less than $2$ " becomes $|A - B| < 2$ , that is $-2 < A - B < 2$ .

Which is larger?

The time Ana takes to run a lap is $X \sim N(30, 9)$ seconds and the time Ben takes is $Y \sim N(28, 16)$ seconds, independently. Find the probability that, in a race, Ana takes longer than Ben.

Solution

Let $D = X - Y$ . Then $E(D) = 30 - 28 = 2$ and $\text{Var}(D) = 9 + 16 = 25$ , so $D \sim N(2, 25)$ .

P(X > Y) = P(D > 0) = P\left(Z > \frac{0 - 2}{5}\right) = P(Z > -0.4) = \Phi(0.4) = 0.655

The shaded area is $P(D > 0)$ for $D \sim N(2, 25)$ :

y = exp(-(x - 2)^2 / 50) / sqrt(50 pi) fill 0 19 y = exp(-(x - 2)^2 / 50) / sqrt(50 pi)

One item against a multiple of another

Large bags of flour have masses $L \sim N(5.2, 0.3^2)$ kg and small bags have masses $S \sim N(1.7, 0.12^2)$ kg. Find the probability that a randomly chosen large bag weighs more than three times a randomly chosen small bag.

Solution

We need $P(L > 3S) = P(L - 3S > 0)$ . Here $3S$ is three times one small bag, so its variance is $3^2 \times 0.12^2$ .

Let $D = L - 3S$ :

E(D) = 5.2 - 3(1.7) = 0.1, \qquad \text{Var}(D) = 0.3^2 + 3^2(0.12^2) = 0.09 + 0.1296 = 0.2196

So $D \sim N(0.1, 0.2196)$ and

P(D > 0) = P\left(Z > \frac{-0.1}{\sqrt{0.2196}}\right) = P(Z > -0.213) = \Phi(0.213) = 0.584

If the question had said "the total of three small bags", the variance would be $3 \times 0.12^2 = 0.0432$ instead.

An exam-style lift question

The masses of men are normally distributed with mean $82$ kg and standard deviation $10$ kg. The masses of women are normally distributed with mean $66$ kg and standard deviation $8$ kg. All masses are independent.

(a) A lift holds $4$ men and $5$ women. Find the probability that their total mass exceeds $700$ kg.

(b) Find the probability that the difference between the masses of two randomly chosen men is less than $2$ kg.

(c) The lift can hold $n$ men. Find the largest value of $n$ for which the probability that the total mass of $n$ randomly chosen men exceeds $1000$ kg is less than $0.05$ .

Solution

(a) Let $T = M_1 + \cdots + M_4 + W_1 + \cdots + W_5$ .

E(T) = 4(82) + 5(66) = 328 + 330 = 658

\text{Var}(T) = 4(10^2) + 5(8^2) = 400 + 320 = 720

$T \sim N(658, 720)$ , so

P(T > 700) = P\left(Z > \frac{700 - 658}{\sqrt{720}}\right) = P(Z > 1.565) = 1 - 0.9412 = 0.0588

(b) Let $D = M_1 - M_2$ . $E(D) = 0$ , $\text{Var}(D) = 100 + 100 = 200$ , so $D \sim N(0, 200)$ .

P(|D| < 2) = P\left(-\frac{2}{\sqrt{200}} < Z < \frac{2}{\sqrt{200}}\right) = P(-0.141 < Z < 0.141)

= 2\Phi(0.141) - 1 = 2(0.5561) - 1 = 0.112

P\left(Z > \frac{1000 - 82n}{10\sqrt{n}}\right) < 0.05 \quad\Rightarrow\quad \frac{1000 - 82n}{10\sqrt{n}} > 1.645

Try values: $n = 11$ gives $\dfrac{98}{33.17} = 2.95 > 1.645$ , which works; $n = 12$ gives $\dfrac{16}{34.64} = 0.462 < 1.645$ , which fails.

The largest value is $n = 11$ .

Solving for n

In (c) the inequality can also be solved as a quadratic in $\sqrt{n}$ : $82n + 16.45\sqrt{n} - 1000 < 0$ . Trial of integer values, clearly shown, is quicker and fully acceptable. Always show the value that works and the next one that fails.

Two-sided events

For "within", "differ by less than" or "between" questions, find the probability for an interval, using symmetry where the mean is $0$ :

P(|D| < k) = 2\Phi\!\left(\frac{k}{\sigma_D}\right) - 1 \quad \text{when } E(D) = 0.

When the mean of $D$ is not zero, work out both $z$ -values separately.

Common mistakes

Using $n^2\sigma^2$ for a total of separate items, or $n\sigma^2$ for a multiple of one item. Read the context: "three bags" is a sum; "three times the mass of a bag" is a multiple.
Subtracting variances for a difference. $\text{Var}(X - Y) = \sigma_1^2 + \sigma_2^2$ .
Comparing two separate probabilities. $P(X > Y)$ is not $P(X > \mu_2)$ or anything built from two single-variable probabilities. Form $X - Y$ .
Mixing up variance and standard deviation. If a question gives standard deviations, square them before combining.
Forgetting the absolute value. "Differ by less than 2" is two-sided: $-2 < X - Y < 2$ .
Missing the order of subtraction. If $D = X - Y$ , then " $Y$ is greater" is $D < 0$ . Write the event in terms of $D$ before standardising.

Exam tip

Write the combination explicitly: " $D = L - 3S$ ", then " $D \sim N(0.1, 0.2196)$ ". Marks are given for the correct mean, the correct variance and the correct final probability.
Show the variance calculation in full: $0.3^2 + 3^2 \times 0.12^2$ . The most common lost mark is a missing square on the coefficient.
Questions may ask you to "state an assumption". The answer is almost always independence, in context: "the masses of the people in the lift are independent".
Keep at least 4 significant figures in the standard deviation before standardising; give $z$ to 3 decimal places.
In multi-stage questions the answer to one part is often used as $p$ in a binomial: "find the probability that at least 2 of 5 lifts are overloaded".

Summary

$aX + b$ is normal if $X$ is normal: $N(a\mu + b, a^2\sigma^2)$ .
$aX + bY$ is normal for independent normals: $N(a\mu_1 + b\mu_2,\ a^2\sigma_1^2 + b^2\sigma_2^2)$ .
Total of $n$ independent items: $N(n\mu, n\sigma^2)$ ; $n$ times one item: $N(n\mu, n^2\sigma^2)$ .
Turn every comparison into a single combination compared with a number, usually $0$ .
Variances add for both sums and differences.
Two-sided events with mean $0$ : $2\Phi(z) - 1$ .

Practice questions

Question

$X \sim N(20, 9)$ and $Y = 5X - 40$ . State the distribution of $Y$ and find $P(Y < 50)$ .
Rods have lengths that are normally distributed with mean $45$ cm and standard deviation $0.4$ cm. Five rods are placed end to end. Find the probability that the total length is between $224$ cm and $226$ cm.
$X \sim N(12, 4)$ and $Y \sim N(10, 5)$ are independent. Find $P(X > Y)$ .
$X \sim N(30, 16)$ and $Y \sim N(19, 4)$ are independent. Find $P(2X > 3Y)$ .
Bolts have diameters $B \sim N(2.00, 0.02^2)$ cm and the holes they must fit have diameters $H \sim N(2.05, 0.03^2)$ cm, independently. A bolt fits a hole if its diameter is smaller than the hole's. Find the probability that a randomly chosen bolt fits a randomly chosen hole.
The mass of coffee in a jar is $N(200, 5^2)$ g and the mass of an empty jar is $N(150, 8^2)$ g, independently. Find the probability that a full jar has total mass more than $360$ g.
Packets of rice have masses $N(500, 15^2)$ g. A box has mass $N(300, 20^2)$ g and holds $24$ packets. Find the probability that a full box weighs more than $12.4$ kg.
Times to run $400$ m are $A \sim N(52, 1.5^2)$ seconds for one runner and $B \sim N(53, 2^2)$ seconds for another, independently. Find the probability that they finish within $1$ second of each other.
$X \sim N(\mu, 9)$ . $X_1$ , $X_2$ and $X_3$ are independent observations of $X$ , and $P(X_1 + X_2 + X_3 > 60) = 0.1$ . Find $\mu$ .
The volume of drink dispensed by a machine into a cup is $V \sim N(250, 6^2)$ ml. Cups have capacity $C \sim N(260, 4^2)$ ml, independently of $V$ . (a) Find the probability that a cup overflows. (b) A customer buys $3$ drinks. Find the probability that the total volume of drink is more than $760$ ml. (c) A smaller machine dispenses volumes $S \sim N(120, 3^2)$ ml. Find the probability that a drink from the large machine is more than twice the volume of a drink from the small machine.

Answers

$E(Y) = 5(20) - 40 = 60$ , $\text{Var}(Y) = 25 \times 9 = 225$ , so $Y \sim N(60, 225)$ . $P(Y < 50) = P\left(Z < \dfrac{-10}{15}\right) = P(Z < -0.667) = 1 - 0.7476 = 0.252$ .
$T \sim N(225, 5 \times 0.16) = N(225, 0.8)$ , $\sigma_T = 0.8944$ . $P(224 < T < 226) = P(-1.118 < Z < 1.118) = 2\Phi(1.118) - 1 = 2(0.8682) - 1 = 0.736$ .
$D = X - Y \sim N(2, 9)$ . $P(D > 0) = P(Z > -0.667) = \Phi(0.667) = 0.748$ .
$D = 2X - 3Y$ : $E(D) = 60 - 57 = 3$ , $\text{Var}(D) = 4(16) + 9(4) = 100$ . $P(D > 0) = P(Z > -0.3) = \Phi(0.3) = 0.618$ .
$D = H - B \sim N(0.05, 0.03^2 + 0.02^2) = N(0.05, 0.0013)$ , $\sigma_D = 0.03606$ . $P(D > 0) = P(Z > -1.387) = \Phi(1.387) = 0.917$ .
$T \sim N(350, 25 + 64) = N(350, 89)$ . $P(T > 360) = P(Z > 1.060) = 1 - 0.8554 = 0.145$ .
$E = 24(500) + 300 = 12\,300$ ; $\text{Var} = 24(225) + 400 = 5800$ ; $\sigma = 76.16$ . $P(T > 12\,400) = P(Z > 1.313) = 1 - 0.9054 = 0.0946$ .
$D = A - B \sim N(-1, 2.25 + 4) = N(-1, 6.25)$ , $\sigma_D = 2.5$ . $P(-1 < D < 1) = P\left(\dfrac{-1 + 1}{2.5} < Z < \dfrac{1 + 1}{2.5}\right) = P(0 < Z < 0.8) = 0.7881 - 0.5 = 0.288$ .
$X_1 + X_2 + X_3 \sim N(3\mu, 27)$ . $\dfrac{60 - 3\mu}{\sqrt{27}} = 1.282$ , so $3\mu = 60 - 1.282 \times 5.196 = 53.339$ and $\mu = 17.8$ (3 s.f.).
(a) Overflow when $V > C$ : $D = V - C \sim N(-10, 36 + 16) = N(-10, 52)$ . $P(D > 0) = P\left(Z > \dfrac{10}{\sqrt{52}}\right) = P(Z > 1.387) = 1 - 0.9173 = 0.0827$ . (b) $T = V_1 + V_2 + V_3 \sim N(750, 108)$ . $P(T > 760) = P\left(Z > \dfrac{10}{\sqrt{108}}\right) = P(Z > 0.962) = 1 - 0.8320 = 0.168$ . (c) $D = V - 2S$ : $E(D) = 250 - 240 = 10$ , $\text{Var}(D) = 36 + 4(9) = 72$ . $P(D > 0) = P\left(Z > \dfrac{-10}{\sqrt{72}}\right) = P(Z > -1.179) = \Phi(1.179) = 0.881$ .

The key facts

Totals

Comparisons: turn them into differences

Two-sided events

Practice questions

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