Mathematics 9709/63 — May/June 2021

Cambridge A-Level · Probability & Statistics 2 · worked solutions for every part, with the mark scheme

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Topics The Poisson Distribution · Linear Combinations of Random Variables · Hypothesis Tests · Sampling and Estimation · Continuous Random Variables

Q15MThe Poisson DistributionLinear Combinations of Random VariablesFree sample

The number of goals scored by a team in a match is independent of other matches, and is denoted by the random variable $X$ , which has a Poisson distribution with mean 1.36. A supporter offers to make a donation of $5 to the team for each goal that they score in the next 10 matches.

Find the expectation and standard deviation of the amount that the supporter will pay.

DifficultyMedium

Worked solution

Approach

Let $T$ be the total number of goals in the next 10 matches. Since each match has $X \sim \text{Po}(1.36)$ and matches are independent, $T$ is Poisson with mean $10 \times 1.36 = 13.6$ . The amount donated is $A = 5T$ . Use linearity of expectation and the variance scaling rule.

Working

\lambda_T = 10 \times 1.36 = 13.6

So $T \sim \text{Po}(13.6)$ , and therefore

\text{E}(T) = 13.6, \quad \text{Var}(T) = 13.6

For the amount $A = 5T$ :

\text{E}(A) = 5 \times \text{E}(T) = 5 \times 13.6 = 68

\text{Var}(A) = 5^2 \times \text{Var}(T) = 25 \times 13.6 = 340

Thus the standard deviation is

\text{SD}(A) = \sqrt{340} = 18.439\ldots = 18.4 \quad (3\text{ s.f.})

Answer

The expected donation is $68 and the standard deviation is $18.4 (3 s.f.).

Final answer

E(amount) = $68; SD(amount) = $18.4 (3 s.f.)

Detailed explanation

Walkthrough

We are told that the number of goals in each match has a Poisson distribution with mean 1.36, and that matches are independent. Over 10 matches, the total number of goals $T$ is the sum of 10 independent Poisson variables. A key property is that the sum of independent Poisson variables is also Poisson, with mean equal to the sum of the means. Since each match has mean 1.36, $T \sim \text{Po}(13.6)$ .

For a Poisson distribution, the variance is equal to the mean, so $\text{Var}(T) = 13.6$ as well. The donation amount is $A = 5T$ . The expectation scales linearly: $\text{E}(A) = 5\text{E}(T) = 5 \times 13.6 = 68$ . The variance, however, scales by the square of the multiplier: $\text{Var}(A) = 5^2 \times \text{Var}(T) = 25 \times 13.6 = 340$ . Finally, the standard deviation is the square root of the variance: $\text{SD}(A) = \sqrt{340} = 18.4$ (3 s.f.).

Key Takeaways

The sum of independent Poisson variables is Poisson, with mean equal to the sum of the means.
For a Poisson distribution, the mean and variance are both equal to $\lambda$ .
If $A = aT$ , then $\text{E}(A) = a\text{E}(T)$ and $\text{Var}(A) = a^2\text{Var}(T)$ ; the standard deviation is $|a|\text{SD}(T)$ .
Always take the square root when a question asks for standard deviation, not variance.

Common Mistakes

Using $\lambda = 1.36$ instead of $\lambda = 13.6$ for the 10 matches.
Writing $\text{Var}(5T) = 5\text{Var}(T)$ instead of $\text{Var}(5T) = 25\text{Var}(T)$ .
Giving the variance 340 as the standard deviation.
Not showing the multiplication by 5 and by $5^2$ ; the mark scheme awards method marks for $\lambda = 10 \times 1.36$ and for $5^2 \times \lambda$ .

Things to Be Careful About

The donation amount is in dollars, so the expectation and standard deviation are dollar amounts.
The final standard deviation should be given to 3 significant figures: 18.4. The exact form $2\sqrt{85}$ is also acceptable.
Do not confuse variance with standard deviation; take the square root at the end.
The mark scheme condones $2\sqrt{85}$ as an exact equivalent, but a rounded answer must be 18.4.

Techniques used

model total goals as a Poisson variable with scaled meanapply linear scaling to expectation and variancetake the square root of variance for standard deviation

The rest of this paper

5 more questions

Q2Hypothesis Tests8M
Q3Hypothesis Tests · The Poisson Distribution · Linear Combinations of Random Variables6M
Q4Sampling and Estimation9M
Q5The Poisson Distribution9M
Q6Continuous Random Variables13M

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