| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Custom discrete distribution sample mean |
| Difficulty | Moderate -0.8 This is a straightforward application of the Central Limit Theorem to a binomial distribution. Part (a) requires calculating mean and variance of X, then applying CLT formulas (dividing variance by n). Part (b) is a routine normal probability calculation with continuity correction. The question involves only direct formula application with no conceptual challenges or problem-solving insight required. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks |
|---|---|
| 5(a) | For X, μ = 2 σ2 = 1.6 |
| Mean = 2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 160 100 | B1 | Accept Var = 0.12 (accept sd=0.1 if clearly identified). |
| Normal | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(b) | 1.8− 1 −'2' 1.8−'2' |
| Answer | Marks | Guidance |
|---|---|---|
| or ± (287.5 – ’320’) / '256' or ± (288 – ‘320’) / '256' [= –2.03 or –2] | M1 | Allow with wrong continuity correction. |
| Answer | Marks | Guidance |
|---|---|---|
| ɸ(‘–2.03’) = 1 – ɸ(‘2.03’) | M1 | Correct area consistent with their values. |
| Answer | Marks |
|---|---|
| = 0.0212 or 0.0228 (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | For X, μ = 2 σ2 = 1.6
Mean = 2 | B1
1.6 1
Variance= or or 0.01
160 100 | B1 | Accept Var = 0.12 (accept sd=0.1 if clearly identified).
Normal | B1
3
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | 1.8− 1 −'2' 1.8−'2'
320 or [= –2.03 or –2]
'0.01' '0.01'
or ± (287.5 – ’320’) / '256' or ± (288 – ‘320’) / '256' [= –2.03 or –2] | M1 | Allow with wrong continuity correction.
M1 can be implied by correct final answer or for
–2.03 / –2.0 or 0.9788 / 0.9772 seen.
ɸ(‘–2.03’) = 1 – ɸ(‘2.03’) | M1 | Correct area consistent with their values.
M1 can be implied by correct final answer.
= 0.0212 or 0.0228 (3 sf) | A1
3
Question | Answer | Marks | Guidance$X$ is a random variable with distribution B(10, 0.2). A random sample of 160 values of $X$ is taken.
\begin{enumerate}[label=(\alph*)]
\item Find the approximate distribution of the sample mean, including the values of the parameters. [3]
\item Hence find the probability that the sample mean is less than 1.8. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q5 [6]}}