Central Limit Theorem

Topic

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

A2 June 2024 Q4

EdexcelCurrent spec12 marksIncludes hypothesis testingCentral Limit Theorem Geometric & Negative Binomial

4. Every morning Geethaka repeatedly rolls a fair, six-sided die until he rolls a 3 and then he stops. The random variable \(X\) represents the number of times he rolls the die each morning.

(a) Suggest a suitable model for the random variable \(X\) (1)

(b) Show that \(\mathrm{P}(X \leqslant 3) = \dfrac{91}{216}\) (2)

After 64 mornings Geethaka will calculate the mean number of times he rolled the die.

Nira wants to check Geethaka’s die to decide whether or not the probability of rolling a 3 with his die is less than \(\dfrac{1}{6}\)

Nira rolls the die repeatedly until she rolls a 3
She obtains \(x = 16\)

(d) By carrying out a suitable test, determine what Nira’s conclusion should be. You should state your hypotheses clearly and use a 5% level of significance. (4)

Mark scheme (a)

Scheme	Marks	AO
\(X \sim \mathrm{Geo}\left(\dfrac{1}{6}\right)\) accept in words: geometric distribution with \(p = \dfrac{1}{6}\)	B1	3.3
	(1)

Notes

B1 for both “geometric” or “Geo” and correct parameter of \(\dfrac{1}{6}\). Probability must be seen in (a)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{P}(X \leqslant 3) = 1 - \mathrm{P}(X \gt 3)\) or \(\mathrm{P}(X = 1) + \mathrm{P}(X = 2) + \mathrm{P}(X = 3)\)	M1	1.1b
\(= 1 - \left(\dfrac{5}{6}\right)^3\) or \(\dfrac{1}{6} + \dfrac{5}{6} \times \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^2 \times \dfrac{1}{6} = \dfrac{91}{216}\) (*)	A1cso	1.1b
	(2)

Notes

M1 for a correct method, may be implied by a correct expression

A1cso* for a correct solution leading to printed answer with no incorrect working seen

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{E}(X) = 6 \qquad \mathrm{Var}(X) = \dfrac{\frac{5}{6}}{\left(\frac{1}{6}\right)^2}\ [= 30]\)	M1 A1	3.4 1.1b
\(\overline{X} \approx\ \sim \mathrm{N}\left(6, \left(\sqrt{\dfrac{30}{64}}\right)^2\right)\)	M1 A1	3.3 1.1b
\(\mathrm{P}(5.6 \lt \overline{X} \lt 7.2) = 0.68064\ldots\)	A1	1.1b
	(5)

Notes

1^st M1 for correct use of formula for \(\mathrm{E}(X)\) or \(\mathrm{Var}(X)\)

1^st A1 for both correct must see value of 6 for \(\mathrm{E}(X)\) and at least correct formula used for \(\mathrm{Var}(X)\) the value of 30 may be implied by 2^nd A1

2^nd M1 for use of CLT to get a normal distribution with correct mean and variance \(\neq\) their 30
Condone use of \(X\) (or their letter) instead of \(\overline{X}\) or seeing \(\mathrm{N}\left(6, \sqrt{\dfrac{30}{64}}\right)\)

2^nd A1 for a correct normal distribution stated or used. May be implied by correct answer.

3^rd A1 for awrt 0.681 but allow 0.68 or 0.680 if correct distribution is seen

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{H}_0: p = \dfrac{1}{6} \qquad \mathrm{H}_1: p \lt \dfrac{1}{6}\)	B1	2.5
Probability approach: \(\mathrm{P}(X \geqslant 16)\) CR approach: \(\left(\dfrac{5}{6}\right)^{c - 1} \lt 0.05\) or \(\left(\dfrac{5}{6}\right)^d \lt 0.05\) with use of logs	M1	3.4
Probability approach: \(= \mathrm{P}(X \gt 15) = \left(\dfrac{5}{6}\right)^{15} = 0.0649\ldots\) CR approach: CR: \(X \geqslant 18\) or \(X \gt 17\)	A1	1.1b
(Not significant) insufficient evidence that probability is \(\lt \dfrac{1}{6}\) or (Not significant) insufficient evidence that dice is biased	A1	2.2b
	(4)
	(12 marks)

Notes

B1 for both hypotheses correct in terms of \(p\)

M1 for sight or use of \(\mathrm{P}(X \geqslant 16)\) (may be implied)
or correct expression with logs (use of logs may be implied by 17.43…)

1^st A1 for 0.065 or better but accept 0.06 if a correct expression is seen, or correct CR

2^nd A1 for a correct conclusion mentioning probability or biased
Condone ‘evidence suggests probability is equal to \(\dfrac{1}{6}\)’
NB: Must have correct probability or correct CR to gain final A1.
NB: Send responses comparing 0.935 with 0.95 to review

A2 June 2023 Q4

EdexcelCurrent spec6 marksCentral Limit Theorem Geometric & Negative Binomial

4. There are 32 students in a class.
Each student rolls a fair die repeatedly, stopping when their total number of sixes is 4
Each student records the total number of times they rolled the die.

Estimate the probability that the mean number of rolls for the class is less than 27.2 (6)

Mark scheme

Scheme	Marks	AO
[\(X\) = no. of rolls to 4 sixes] \(\quad X \sim \mathrm{NegBin}\left(4, \tfrac{1}{6}\right)\)	M1	3.3
\(\mu\left[= \dfrac{r}{p}\right] = \underline{\mathbf{24}}, \quad \sigma^2\left[= \dfrac{r(1 - p)}{p^2}\right] = \dfrac{4 \times \frac{5}{6}}{\frac{1}{36}} = \underline{\mathbf{120}}\)	A1, A1	1.1b(x2)
[\(\overline{X} \approx\sim\)] \(\mathrm{N}\left(\text{“}24\text{”}, \sqrt{\dfrac{\text{“}120\text{”}}{32}}^{\,2}\right)\)	M1 M1	2.1,3.4
\(\mathrm{P}(\overline{X} \lt 27.2) = 0.95078\ldots\) awrt 0.951	A1	1.1b
	(6 marks)

Notes

1^st M1 for selecting the correct negative binomial model. May be implied by correct mean or variance
NegBin on its own is M0

1^st A1 for mean = 24

2^nd A1 for variance = 120 \(\sigma = 120\) is A0 unless recovered

2^nd M1 for writing or using of normal with mean 24 (may be implied by correct answer)
ft their mean which may come from any distribution

3^rd M1 for writing or using normal with standard deviation \(= \sqrt{\dfrac{120}{32}}\ \left[= \sqrt{3.75}\right]\)
ft \(\dfrac{\text{their } \sigma}{\sqrt{32}}\) where \(\sigma\) may come from any distribution
(may be implied by correct answer)

2^nd A1 for awrt 0.951 (correct answer scores 6 out of 6)

A2 June 2022 Q5

EdexcelCurrent spec5 marksCentral Limit Theorem Geometric & Negative Binomial

5. A random sample of 150 observations is taken from a geometric distribution with parameter 0.3

Estimate the probability that the mean of the sample is less than 3.45 (5)

Mark scheme

Scheme	Marks	AO
\(\mathrm{Geo}(0.3) \quad \mu = \dfrac{1}{0.3}\left[\text{or exact equivalent e.g. } \dfrac{10}{3}\right]\)	B1	1.1b
\(\sigma^2 = \dfrac{1 - 0.3}{0.3^2}\left[\text{or exact equivalent e.g. } \dfrac{70}{9}\right]\)	B1	1.1b
CLT \(\Rightarrow \overline{X} \approx \mathrm{N}\left(\dfrac{10}{3}, \ldots\right)\) oe	M1	2.1
\(\Rightarrow \overline{X} \approx \mathrm{N}\left(\dfrac{10}{3}, \dfrac{7}{135}\right)\) and attempt (sight of) \(\mathrm{P}(\overline{X} \lt 3.45)\)	M1	3.4
\(= 0.69579\ldots\) awrt 0.696	A1	1.1b
	(5)
	Total 5

Notes

1^st B1 correct mean

2^nd B1 correct Var may be implied by sight of \(\frac{7}{135}\) in distribution of \(\overline{X}\)

1^st M1 For use of CLT (must see \(\overline{X}\) and Normal with mean correct ft) or sight of \(\mathrm{N}\left(\dfrac{10}{3}, \dfrac{7}{135}\right)\) or \(\mathrm{N}\left(\text{“}\dfrac{10}{3}\text{”}, \dfrac{\text{“}70\text{”}}{\text{“}9\text{”} \times 150}\right)\) with any letter
Allow 3.33 or better for \(\frac{10}{3}\) and 7.78 or better for \(\frac{70}{9}\)
May be implied by 2^nd M1

2^nd M1 Using the normal distribution to find \(\mathrm{P}(\overline{X} \lt 3.45)\) ft their \(\text{“}\dfrac{10}{3}\text{”}\) and \(\dfrac{\text{“}\dfrac{70}{9}\text{”}}{150}\)
May be implied by correct answer.

A1 awrt 0.696
Correct answer with no incorrect working scores 5/5

Alternative (Use of \(Y = \sum X\))

Scheme	Marks
\(\mu = \dfrac{150}{0.3}\ [= 500]\)	B1
\(\sigma^2 = \dfrac{150 \times 0.7}{0.3^2}\left[\dfrac{3500}{3}\right] = 1166.\dot{6}\)	B1
\(\Rightarrow Y \approx \mathrm{N}\left(500, \dfrac{3500}{3}\right)\)	M1
\(\mathrm{P}(Y \lt 517.5)\)	M1
\(= 0.69579\ldots\)	A1

A2 October 2021 Q3

EdexcelCurrent spec4 marksCentral Limit Theorem

3. A courier delivers parcels. The random variable \(X\) represents the number of parcels delivered successfully each day by the courier where \(X \sim \mathrm{B}(400, 0.64)\)

A random sample \(X_1, X_2, \ldots X_{100}\) is taken.

Estimate the probability that the mean number of parcels delivered each day by the courier is greater than 257 (4)

Mark scheme

Scheme	Marks	AO
\(\overline{X} \approx \mathrm{N}(256, \ldots)\) oe	M1	3.1a
\(\overline{X} \approx \mathrm{N}(256, 0.9216)\)	A1	1.1b
\(\mathrm{P}(\overline{X} \gt 257) = \mathrm{P}\left(Z \gt \dfrac{257 - 256}{\sqrt{\text{“}0.9216\text{”}}}\right)\ [= \text{awrt } 1.04]\)	dM1	3.4
\(p = 0.1492\ldots\)	A1	1.1b
	(4)
	(4 marks)

Notes

M1: For realising the need to use the CLT with correct mean

A1: For a correct normal stated

dM1: Dep on previous Method mark. Use of the normal model to find \(\mathrm{P}(\overline{X} \gt 257)\) If final answer is incorrect then we need to see the standardisation using their \(\sigma\).

A1: awrt 0.149 (0.14878… from calculator)

NB Allow awrt 0.148 if a continuity correction is used.

A2 October 2020 Q7

EdexcelCurrent spec15 marksIncludes hypothesis testingCentral Limit Theorem Quality of Tests

7. A six-sided die has sides labelled 1, 2, 3, 4, 5 and 6

The random variable \(S\) represents the score when the die is rolled.

Alicia rolls the die 45 times and the mean score, \(\overline{S}\), is calculated.

Assuming the die is fair and using a suitable approximation,

(a) find, to 3 significant figures, the value of \(k\) such that \(\mathrm{P}(\overline{S} \lt k) = 0.05\) (8)

(b) Explain the relevance of the Central Limit Theorem in part (a). (2)

Alicia considers the following hypotheses:

\(\mathrm{H}_0\): The die is fair

\(\mathrm{H}_1\): The die is not fair

If \(\overline{S} \lt 3.1\) or \(\overline{S} \gt 3.9\), then \(\mathrm{H}_0\) will be rejected.

Given that the true distribution of \(S\) has mean 4 and variance 3

(d) Describe what would happen to the power of this test if Alicia were to increase the number of rolls of the die.
Give a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks	AO
Realising \(S\) has a discrete uniform distribution over \(\{1, \ldots 6\}\)	M1	3.3
\(\mathrm{E}(S) = 1 \times \tfrac{1}{6} + 2 \times \tfrac{1}{6} + 3 \times \tfrac{1}{6} + 4 \times \tfrac{1}{6} + 5 \times \tfrac{1}{6} + 6 \times \tfrac{1}{6}\)	M1	1.1b
\(\mathrm{Var}(S) = \tfrac{6^2 - 1}{12}\) or \(1^2 \times \tfrac{1}{6} + 2^2 \times \tfrac{1}{6} + 3^2 \times \tfrac{1}{6} + 4^2 \times \tfrac{1}{6} + 5^2 \times \tfrac{1}{6} + 6^2 \times \tfrac{1}{6} - 3.5^2\)	M1	1.1b
\(\mathrm{E}(S) = 3.5\) and \(\mathrm{Var}(S) = \tfrac{35}{12}\)	A1	1.1b
\(\overline{S} \sim \mathrm{N}(3.5, \ldots)\)	M1	3.1a
\(\mathrm{Var}(\overline{S}) = \dfrac{\frac{35}{12}}{45} = \tfrac{7}{108}, \ \ \overline{S} \sim \mathrm{N}(3.5, 0.0648\ldots)\)	A1	1.1b
\(\mathrm{P}(\overline{S} \lt k) = 0.05 \rightarrow \dfrac{k - 3.5}{\sqrt{\frac{7}{108}}} = -1.6449\)	M1	3.4
\(k = 3.08122\ldots\) awrt 3.08	A1	1.1b
	(8)

Notes

M1: Setting up model for \(S\)

M1: Attempt at expression for \(\mathrm{E}(S)\)

M1: Attempt at expression for \(\mathrm{Var}(S)\)

A1: Correct mean and variance for \(S\) (may be implied by a correct distribution for \(\overline{S}\))

M1: Use of CLT to find distribution for \(\overline{S} \sim \mathrm{N}(\text{‘}3.5\text{’}, \ldots)\) f.t. their 3.5 but variance \(\neq \tfrac{35}{12}\)

A1: Correct distribution with correct variance, allow \(\sigma^2 =\) awrt 0.0648 or \(\sigma =\) awrt 0.255

M1: Standardising using their model and equating to a \(z\)-value \(1 \lt |z| \lt 2\)

A1: awrt 3.08

Mark scheme (b)

Scheme	Marks	AO
CLT applies since the sample size is large	B1	3.5b
CLT states that the sample mean/\(\overline{S}\) is (approximately) normally distributed	B1	3.5b
	(2)

Notes

B1: Correct explanation about appropriateness of the CLT given large sample size (allow > 30)

B1: Requires both sample and mean or \(\overline{S}\)

Mark scheme (c)

Scheme	Marks	AO
True \(\overline{S} \sim \mathrm{N}(4, \tfrac{3}{45})\)	M1	3.3
\(\mathrm{P}(\overline{S} \lt 3.1) + \mathrm{P}(\overline{S} \gt 3.9)\) or \(1 - \mathrm{P}(3.1 \lt \overline{S} \lt 3.9)\)	dM1	3.4
Power = awrt 0.651	A1	1.1b
	(3)

Notes

M1: Writing or using \(\overline{S} \sim \mathrm{N}(4, \tfrac{3}{45})\) allow \(\sigma^2 =\) awrt 0.0667 or \(\sigma =\) awrt 0.258

dM1: (dep on 1^st M1) correct probability statement for power

A1: awrt 0.651

Mark scheme (d)

Scheme	Marks	AO
E.g. The increase in sample size would decrease the variance of \(\overline{S}\) [leading to an increase in \(\mathrm{P}(\overline{S} \gt 3.9)\) and the decrease in \(\mathrm{P}(\overline{S} \lt 3.1)\) would be negligible]	B1	2.4
So the power would increase.	dB1	2.2a
	(2)
	(15 marks)

Notes

B1: Correct reasoning which refers to decrease in variance

dB1: (dep on 1^st B1) Correct deduction with no incorrect reasoning

A2 June 2019 Q3

EdexcelCurrent spec6 marksCentral Limit Theorem DRVs

3. A biased spinner can land on the numbers 1, 2, 3, 4 or 5 with the following probabilities.

Number on spinner	1	2	3	4	5
Probability	0.3	0.1	0.2	0.1	0.3

The spinner will be spun 80 times and the mean of the numbers it lands on will be calculated.

Find an estimate of the probability that this mean will be greater than 3.25 (6)

Mark scheme

Scheme	Marks	AO
{Let \(X\) = the number when the spinner is spun} \(\mu = \underline{3}\)	B1	1.1b
\([\mathrm{E}(X^2) =]\ 0.3 + 4 \times 0.1 + 9 \times 0.2 + 16 \times 0.1 + 25 \times 0.3 \quad \left[= 11.6 \text{ or } \tfrac{58}{5}\right]\)	M1	1.1b
\(\sigma^2\ [= 11.6 - 3^2 =]\ \underline{2.6}\)	A1	1.1b
\(\overline{X} \mathrel{\approx\!\sim} \mathrm{N}\left(\text{“}3\text{”}, \sqrt{\dfrac{\text{“}2.6\text{”}}{80}}^{\,2}\right)\)	M1 A1ft	2.1 1.1b
\(\mathrm{P}(\overline{X} \gt 3.25) = [\mathrm{P}(Z \gt 1.3867\ldots) =]\ 0.0827589\ldots\) (calc) awrt 0.0828	A1	3.4
	(6 marks)

Notes

B1 for stating or using mean = 3

1^st M1 for using the given model to attempt \(\mathrm{E}(X^2)\) with at least 3 correct products seen

1^st A1 for \(\mathrm{Var}(X) = 2.6\) or \(\sigma = \sqrt{2.6} = 1.6124\ldots\) (awrt 1.61)

ALT Use of pgf (B1 when mean = 3 seen) (M1 when correct \(\mathrm{G}^{\prime\prime}(t)\) seen with attempt at \(\mathrm{G}^{\prime\prime}(1)\))

\[\begin{aligned}\mathrm{G}(t) &= 0.3t + 0.1t^2 + 0.2t^3 + 0.1t^4 + 0.3t^5\\ \mathrm{G}^{\prime}(t) &= 0.3 + 0.2t + 0.6t^2 + 0.4t^3 + 1.5t^4\\ \mathrm{G}^{\prime\prime}(t) &= 0.2 + 1.2t + 1.2t^2 + 6t^3 \quad \text{leading to } \mathrm{G}^{\prime\prime}(1) = 8.6\end{aligned}\]

2^nd M1 for use of CLT – must use \(\overline{X}\) and normal or sight of \(\mathrm{N}\left(\text{“}3\text{”}, \sqrt{\dfrac{\text{“}2.6\text{”}}{80}}^{\,2}\right)\) with any letter

2^nd A1ft for a correct mean and variance, ft their 3 and their 2.6
This M1A1ft may be implied by sight of correct st. dev. used in a standardisation leading to \(\mathrm{P}(Z \gt 1.39)\) Must see correct use of \(Z\)
NB \(\dfrac{2.6}{80} = 0.0325\) and \(\sqrt{\dfrac{2.6}{80}} = 0.18027\ldots\) so allow e.g. \(\mathrm{N}(3, \text{awrt } (0.180)^2)\)

3^rd A1 for using the normal model to find probability awrt 0.0828

ALT Use of \(\sum X\) (If see clear attempt at \(\mathrm{P}(\Sigma X \gt 260)\) condone \(\mathrm{P}(\Sigma X \gt 260.5)\)) then:

2^nd M1 for \(\Sigma X \sim \mathrm{N}(\ldots)\) or any letter \(\sim \mathrm{N}\left(\text{“}240\text{”}, \sqrt{\text{“}2.6\text{”} \times 80}^{\,2}\right)\)

2^nd A1ft for mean \(= \text{“}3\text{”} \times 80 = 240\) and variance \(= \text{“}2.6\text{”} \times 80 = 208\)

May see \(\mathrm{P}(\Sigma X \gt 260.5) = 0.077597\ldots\) but it will only score 2^nd M1 2^nd A1ft and 3^rd A0

S3 June 2011 Q1

EdexcelOld spec3 marksCentral Limit Theorem

1. Explain what you understand by the Central Limit Theorem. (3)

Mark scheme

Scheme	Marks
\(X_1, X_2, \ldots X_n\) is a random sample of size \(n\), for large \(n\),	B1
drawn from a population of any distribution with mean \(\mu\) and variance \(\sigma^2\)	B1
then \(\overline{X}\) is (approximately) \(\mathrm{N}\left(\mu, \dfrac{\sigma^2}{n}\right)\)	B1
	(3)
	(3 marks)

Notes

1st B for large sample or equivalent

2nd B for ‘population of any distribution’ or ‘any population’

3rd B require mean or symbol and normal ( parameters not required)

S3 June 2009 Q4

EdexcelOld spec5 marksCentral Limit Theorem

4. A sample of size 8 is to be taken from a population that is normally distributed with mean 55 and standard deviation 3. Find the probability that the sample mean will be greater than 57. (5)

Mark scheme

Scheme	Marks
\(X \sim \mathrm{N}(55, 3^2)\) therefore \(\overline{X} \sim \mathrm{N}\left(55, \dfrac{9}{8}\right)\)	B1 B1
\(\mathrm{P}(\overline{X} \gt 57) = \mathrm{P}\left(Z \gt \dfrac{57 - 55}{\sqrt{\frac{9}{8}}}\right) \quad = \mathrm{P}(Z \gt 1.8856\ldots)\)	M1
\(= 1 - 0.9706\)	M1
\(= 0.0294\) 0.0294~0.0297	A1
	(5 marks)

Notes

1st B1 for \(\overline{X}\) ~ normal and \(\mu = 55\), may be implied but must be \(\overline{X}\)

2nd B1 for Var(\(\overline{X}\)) or st. dev of \(\overline{X}\) e.g. \(\overline{X} \sim \mathrm{N}(55, \tfrac{9}{8})\) or \(\overline{X} \sim \mathrm{N}\left(55, \left(\tfrac{3}{\sqrt{8}}\right)^2\right)\) for B1B1
Condone use of \(X\) if they clearly mean \(\overline{X}\) so \(X \sim \mathrm{N}\left(55, \tfrac{9}{8}\right)\) is OK for B1B1

1st M1 for an attempt to standardize with 57 and mean of 55 and their st. dev. \(\ne 3\)

2nd M1 for 1 - tables value. Must be trying to find a probability < 0.5

A1 for answers in the range 0.0294~0.0297

ALT \(\displaystyle\sum_1^8 X_i \sim \mathrm{N}\left(8 \times 55, 8 \times 3^2\right)\)
1st B1 for \(\sum X\) ~ normal and mean \(= 8 \times 55\)
2nd B1 for variance \(= 8 \times 3^2\)
1st M1 for attempt to standardise with \(57 \times 8\), mean of \(55 \times 8\) and their st dev \(\ne 3\)

S3 June 2006 Q2

EdexcelOld spec6 marksCentral Limit Theorem

2. A report on the health and nutrition of a population stated that the mean height of three-year old children is 90 cm and the standard deviation is 5 cm. A sample of 100 three-year old children was chosen from the population.

(a) Write down the approximate distribution of the sample mean height. Give a reason for your answer. (3)

(b) Hence find the probability that the sample mean height is at least 91 cm. (3)

Scheme	Marks
\(\bar{X} \sim \mathrm{N}\left(90, \dfrac{5^2}{100}\right)\) i.e. \(\mathrm{N}(90, 0.25)\)	M1 A1
Application of central limit theorem as (sample large)	B1
	(3)

Mark scheme (b)

Scheme	Marks
\(\mathrm{P}(\bar{X} \geqslant 91) = 1 - \mathrm{P}\left(Z \lt \dfrac{91 - 90}{0.5}\right)\) Stand.	M1 A1
\(= 1 - \mathrm{P}(Z \lt 2)\) \(= 1 - 0.9772\) \(= 0.0228\) awrt 0.0228	A1
	(3)
	(6 marks)

S3 June 2005 Q2

EdexcelOld spec6 marksCentral Limit Theorem

2. A sample of size 5 is taken from a population that is normally distributed with mean 10 and standard deviation 3. Find the probability that the sample mean lies between 7 and 10. (6)

Mark scheme

Scheme	Marks
\(X \sim \mathrm{N}(10, 3^2) \quad \therefore \quad \bar{X} \sim \mathrm{N}\left(10, \tfrac{9}{5}\right)\) Can be implied 10; \(\tfrac{9}{5}\)	B1; B1
\(\mathrm{P}(7 \leqslant \bar{X} \leqslant 10) = \mathrm{P}\left(\dfrac{7 - 10}{\sqrt{9/5}} \lt Z \lt 0\right)\) Standardising with 10 & their \(\sigma\)	M1 A1
\(= \mathrm{P}(-2.236 \lt Z \lt 0)\)
\(= \Phi(0) - \{1 - \Phi(2.24)\}\)	M1 (\(p \lt 0.5\))
\(= \underline{0.4875}\)	A1
	(6)
	(6 marks)

Central Limit Theorem

Edexcel

Edexcel · Old spec

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Alternative (Use of \(Y = \sum X\))

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