Learn9709Paper 6 · Probability & Statistics 2Linear combinations of random variablesCombining normal and Poisson variables

Linear combinations of random variables · Subtopic 3 of 3

Combining Normal and Poisson Variables

Any linear combination of independent normals is normal, so "is $X$ bigger than $Y$ ?" becomes "is $X-Y$ positive?". Poisson variables add to a Poisson, but only sums: $2X$ and $X-Y$ are not Poisson.

Cambridge 9709 Paper 6Edexcel YMA01 WST03Edexcel 9FM0 Option≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find the distribution of $aX+bY$ for independent normal $X$ and $Y$ .
Turn comparisons such as $P(X>Y)$ into $P(X-Y>0)$ .
Handle comparisons with coefficients, such as $P(A>1.4B)$ .
Know when a combination of Poisson variables is Poisson.

Pause and think

Ali's run times are $N(50,9)$ seconds and Bilal's are $N(45,16)$ . Bilal is faster on average. How could you find the probability that Ali wins a race?

Section 2 of 14

Core idea

Start hereIs a randomly chosen man heavier?

Men's weights $N(80,100)$ , women's $N(65,64)$ . Pick one of each: what's the chance the man is heavier? Look at the difference $M-W$ : it's normal too, and the question becomes a single normal probability.

The big idea

Linear combinations of independent normals are normal (use the mean/variance rules). Sums of independent Poissons are Poisson: $\operatorname{Po}(\lambda)+\operatorname{Po}(\mu)=\operatorname{Po}(\lambda+\mu)$ .

1. Normal

Is a man (N(80, 100)) heavier than a woman (N(65, 64))?

Mean

80-65=15

, variance

100+64=164

P(M-W>0)=P\left(Z>\dfrac{-15}{12.81}\right)=P(Z>-1.171)=0.879

Worked exampleMan vs woman

P(man heavier)?

Goal: $P(M-W>0)$ .
Strategy: Distribution of the difference.

$D=M-W\sim N(15,164)$
Means subtract; variances add.
$z=\tfrac{0-15}{12.81}=-1.171$
Standardise.
$P(D>0)=\Phi(1.171)=0.879$
Symmetry.

Answer

0.879

✓ CheckMean difference positive, so above

0.5

✓.

Worked exampleOne man vs two women together

P(one man heavier than two women combined)?

Goal: Probability.
Strategy: $M-(W_1+W_2)$ .

$N(80-130,\ 100+64+64)=N(-50,228)$
Two separate women.
$z=\tfrac{50}{15.1}=3.31$
Standardise.

AnswerAbout

0.0005

✓ CheckVery unlikely: two women average

130

kg vs

80

✓.

2. Poisson

Worked example $X\sim\operatorname{Po}(2)$ , $Y\sim\operatorname{Po}(3)$

$P(X+Y=4)$ ?

Goal: Probability.
Strategy: $X+Y\sim\operatorname{Po}(5)$ .

$e^{-5}\tfrac{5^4}{4!}=e^{-5}\tfrac{625}{24}$
Formula.
$=0.175$
Evaluate.

Answer

0.175

✓ CheckMean of sum

5

2+3

✓.

Common misconceptionNot for differences or multiples

X-Y

can be negative (not Poisson).

3X

has mean

6

but variance

18

(not Poisson).

Pause and think

Is $X_1+X_2$ (two independent Po(4)) Poisson?

Summary

Normal combinations stay normal.
Poisson sums stay Poisson.

Linear combination: $aX+bY+\cdots$

HistoryGauss and errors. Gauss's normal distribution came from studying measurement errors. If each error is normal and errors combine linearly (as when a position is calculated from several measured angles), the final error is still normal, a property that made the distribution so useful in 19th-century astronomy and surveying.

Fun factNormal is "stable". Very few distributions keep their shape when you add independent copies. The normal does, and statisticians call it a "stable" distribution. Add two uniform distributions and you get a triangle, not a uniform; add two normals and you get a normal every time.

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