A2 June 2025 Q5

EdexcelCurrent spec10 marksCentral Limit Theorem Poisson Distribution

5. A football team scores goals at an average rate of 1.8 goals per match.

(a) Give two assumptions that would be necessary to use a Poisson distribution to model the number of goals scored in a match by the team. (2)

Given that in their next match the team scores exactly 3 goals,

(b) find the exact probability that 2 of these goals were scored in the first half of the match. (4)

A hockey team plays 2 games each week during a 40-week season.
Each game consists of 2 periods.
The probability that the team concedes no goals in a period is 0.55

The random variable \(X\) represents the number of periods in a week in which the team concedes no goals.

(c)

(i) Write down a suitable distribution for \(X\)

(ii) For this 40-week season, use the Central Limit Theorem to estimate \(\mathrm{P}(\overline{X} \gt 2)\) (4)

Mark scheme (a)

Scheme	Marks	AO
One correct assumption in context from below	B1	3.5b
Two correct assumptions from below, with goals or matches mentioned at least once Goals are scored at a constant rate Goals are scored independently/randomly or matches are independent Goals are scored not scored simultaneously or Goals are scored singly	B1	3.5b
	(2)

Notes

B1: One contextual correct assumption relating to independence/constant rate/singly
Must mention goals or match, or equivalent context.

B1: Two correct contextual assumptions with goals/matches mentioned at least once.
Assumptions relating to probability or number of goals are B0, but do not penalise if two correct assumptions also given.

Mark scheme (b)

Scheme	Marks	AO
\(A_{0.5} \sim \mathrm{Po}(0.9)\) and \(A_1 \sim \mathrm{Po}(1.8)\)	B1	3.3
\(\mathrm{P}(A_{0.5} = 2 \mid A_1 = 3) = \dfrac{\mathrm{P}(A_{0.5} = 2) \times \mathrm{P}(A_{0.5} = 1)}{\mathrm{P}(A_1 = 3)}\)	M1	3.4
\(\mathrm{P}(A_{0.5} = 2 \mid A_1 = 3) = \dfrac{\dfrac{0.9^2\mathrm{e}^{-0.9}}{2!} \times \dfrac{0.9\mathrm{e}^{-0.9}}{1!}}{\dfrac{1.8^3\mathrm{e}^{-1.8}}{3!}}\) or \(\dfrac{0.164\ldots \times 0.365\ldots}{0.160\ldots}\)	M1	1.1b
\(\mathrm{P}(A_{0.5} = 2 \mid A_1 = 3) = \dfrac{3}{8}\) \(\dfrac{3}{8}\) or exact equivalent	A1	1.1b
	(4)

Notes

B1: Stating or using both correct Poisson distributions [0.164… or 0.365… implies \(\mathrm{Po}(0.9)\), 0.160… implies \(\mathrm{Po}(1.8)\). Both are required for this mark]
or writing or using \(\mathrm{B}(3, 0.5)\)

M1: Correct use of conditional probability formula. Implied by \(\dfrac{0.164\ldots \times 0.365\ldots}{0.160\ldots}\)
Or writing or using any binomial distribution and \(\mathrm{P}(X = 2)\)

M1: Correct expression with probabilities or writing or using \(\mathrm{B}(3, 0.5)\) and \(\mathrm{P}(X = 2)\)

A1: Correct exact answer. Do not award if there is clear evidence of rounding.

Mark scheme (c)

Scheme	Marks	AO
(i) \(X \sim \mathrm{B}(4, 0.55)\)	B1	3.3
(ii) \(\mathrm{E}(X) = \text{‘}4\text{’} \times \text{‘}0.55\text{’}\ [= 2.2]\) or \(\mathrm{Var}(X) = \text{‘}4\text{’} \times \text{‘}0.55\text{’} \times (1 - \text{‘}0.55\text{’})\ [= 0.99]\)	M1	2.1
\(\overline{X} \simeq \mathrm{N}\left(2.2, \dfrac{0.99}{40}\right)\)	M1	3.3
\(\mathrm{P}(\overline{X} \gt 2)\) awrt 0.898	A1	3.4
	(4)
	(10 marks)

Notes

(i) B1: Stating correct binomial distribution

(ii) M1: Correct method to find the mean or variance of \(X\) (a mean of 2.2 from Poisson is M1)
ft their \(n\) and \(p\)

M1: Correct approximate normal distribution [condone using \(X\) not \(\overline{X}\)]. Allow variance awrt 0.0248

A1: awrt 0.898
NB: awrt 0.898 implies the M1M1A1