| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown distribution, CLT applied |
| Difficulty | Standard +0.8 This requires knowledge of the geometric distribution's mean and variance, application of CLT to a non-normal distribution, standardization, and normal table lookup. While mechanically straightforward for Further Maths students, it combines multiple concepts (geometric distribution properties, CLT conditions, z-score calculation) making it moderately above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Geo}(0.3) \quad \mu = \dfrac{1}{0.3}\) [or exact equivalent e.g. \(\dfrac{10}{3}\)] | B1 | Correct mean |
| \(\sigma^2 = \dfrac{1-0.3}{0.3^2}\) [or exact equivalent e.g. \(\dfrac{70}{9}\)] | B1 | Correct Var; may be implied by sight of \(\frac{7}{135}\) in distribution of \(\bar{X}\) |
| \(\text{CLT} \Rightarrow \bar{X} \approx N\!\left(\dfrac{10}{3}, \ldots\right)\) | M1 | For use of CLT (must see \(\bar{X}\) and Normal with mean correct ft) or sight of \(N\!\left(\frac{10}{3}, \frac{7}{135}\right)\) or \(N\!\left("\frac{10}{3}", \frac{"70"}{" 9"\times150}\right)\) with any letter. Allow 3.33 or better for \(\frac{10}{3}\) and 7.78 or better for \(\frac{70}{9}\) |
| \(\Rightarrow \bar{X} \approx N\!\left(\dfrac{10}{3}, \dfrac{7}{135}\right)\) and attempt (sight of) \(P(\bar{X}<3.45)\) | M1 | Using normal distribution to find \(P(\bar{X}<3.45)\) ft their \(\frac{10}{3}\) and \(\frac{70/9}{150}\) |
| \(= 0.69579\ldots\) awrt \(\mathbf{0.696}\) | A1 | Correct answer with no incorrect working scores 5/5 |
## Question 5:
| $\text{Geo}(0.3) \quad \mu = \dfrac{1}{0.3}$ [or exact equivalent e.g. $\dfrac{10}{3}$] | B1 | Correct mean |
| $\sigma^2 = \dfrac{1-0.3}{0.3^2}$ [or exact equivalent e.g. $\dfrac{70}{9}$] | B1 | Correct Var; may be implied by sight of $\frac{7}{135}$ in distribution of $\bar{X}$ |
| $\text{CLT} \Rightarrow \bar{X} \approx N\!\left(\dfrac{10}{3}, \ldots\right)$ | M1 | For use of CLT (must see $\bar{X}$ and Normal with mean correct ft) or sight of $N\!\left(\frac{10}{3}, \frac{7}{135}\right)$ or $N\!\left("\frac{10}{3}", \frac{"70"}{" 9"\times150}\right)$ with any letter. Allow 3.33 or better for $\frac{10}{3}$ and 7.78 or better for $\frac{70}{9}$ |
| $\Rightarrow \bar{X} \approx N\!\left(\dfrac{10}{3}, \dfrac{7}{135}\right)$ and attempt (sight of) $P(\bar{X}<3.45)$ | M1 | Using normal distribution to find $P(\bar{X}<3.45)$ ft their $\frac{10}{3}$ and $\frac{70/9}{150}$ |
| $= 0.69579\ldots$ awrt $\mathbf{0.696}$ | A1 | Correct answer with no incorrect working scores 5/5 |
---\begin{enumerate}
\item A random sample of 150 observations is taken from a geometric distribution with parameter 0.3
\end{enumerate}
Estimate the probability that the mean of the sample is less than 3.45
\hfill \mbox{\textit{Edexcel FS1 2022 Q5 [5]}}