| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Expectation and variance of Poisson-related expressions |
| Difficulty | Standard +0.8 This S4 question requires understanding of Poisson distribution properties, unbiased estimators, and variance calculations across multiple parts. While the individual steps (finding k from unbiasedness, computing variances, comparing estimators) are standard techniques for Further Maths Statistics, the multi-part structure with different sample sizes and the comparison of estimator efficiency in part (e) requires careful algebraic manipulation and conceptual understanding beyond routine application. It's moderately challenging for S4 level but follows predictable patterns. |
| Spec | 5.02i Poisson distribution: random events model5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X_1 \sim Po(3\lambda)\), \(X_2 \sim Po(7\lambda)\), \(X_3 \sim Po(10\lambda)\) | M1 | M1 all 3 needed, Poisson and mean |
| \(E(\hat{\lambda}) = k[E(X_1) + E(X_2) + E(X_3)] = 20\lambda k\) | M1 | M1 adding their means |
| \(\hat{\lambda}\) unbiased therefore \(20\lambda k = \lambda\) | M1 | M1 putting \(E(\hat{\lambda}) = \lambda\) |
| \(k = \frac{1}{20}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(\hat{\lambda}) = \frac{1}{20^2}\text{Var}(X_1 + X_2 + X_3)\) | M1 | M1 use of \(k^2\text{Var}(X_1+X_2+X_3)\) |
| \(= \frac{1}{20^2}(3\lambda + 7\lambda + 10\lambda)\) | M1 | M1 using their means from (a) as variances and adding |
| \(= \frac{\lambda}{20}\) | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim Po(4\lambda)\) | ||
| \(E\!\left(\frac{1}{4}\bar{Y}\right) = \frac{1}{4} \times 4\lambda = \lambda\), therefore unbiased | M1 A1 | M1 use of \(4\lambda\); A1 cso plus conclusion; accept working out bias \(= 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}\!\left(\frac{1}{4}\bar{Y}\right) = \frac{1}{16} \times \frac{4\lambda}{n}\) | M1 B1 | M1 for \(\frac{1}{16}\times\text{Var}\bar{Y}\); B1 for \(\text{Var}\bar{Y} = \frac{4\lambda}{n}\) |
| \(= \frac{\lambda}{4n}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\lambda}{4n} < \frac{\lambda}{20}\) | M1 | M1 for \(\text{Var}\!\left(\frac{1}{4}\bar{Y}\right) < \text{Var}(\hat{\lambda})\) |
| \(n > 5\) therefore \(n = 6\) | A1 |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X_1 \sim Po(3\lambda)$, $X_2 \sim Po(7\lambda)$, $X_3 \sim Po(10\lambda)$ | M1 | M1 all 3 needed, Poisson and mean |
| $E(\hat{\lambda}) = k[E(X_1) + E(X_2) + E(X_3)] = 20\lambda k$ | M1 | M1 adding their means |
| $\hat{\lambda}$ unbiased therefore $20\lambda k = \lambda$ | M1 | M1 putting $E(\hat{\lambda}) = \lambda$ |
| $k = \frac{1}{20}$ | A1 | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(\hat{\lambda}) = \frac{1}{20^2}\text{Var}(X_1 + X_2 + X_3)$ | M1 | M1 use of $k^2\text{Var}(X_1+X_2+X_3)$ |
| $= \frac{1}{20^2}(3\lambda + 7\lambda + 10\lambda)$ | M1 | M1 using their means from (a) as variances and adding |
| $= \frac{\lambda}{20}$ | A1ft | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim Po(4\lambda)$ | | |
| $E\!\left(\frac{1}{4}\bar{Y}\right) = \frac{1}{4} \times 4\lambda = \lambda$, therefore unbiased | M1 A1 | M1 use of $4\lambda$; A1 cso plus conclusion; accept working out bias $= 0$ |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}\!\left(\frac{1}{4}\bar{Y}\right) = \frac{1}{16} \times \frac{4\lambda}{n}$ | M1 B1 | M1 for $\frac{1}{16}\times\text{Var}\bar{Y}$; B1 for $\text{Var}\bar{Y} = \frac{4\lambda}{n}$ |
| $= \frac{\lambda}{4n}$ | A1 | |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\lambda}{4n} < \frac{\lambda}{20}$ | M1 | M1 for $\text{Var}\!\left(\frac{1}{4}\bar{Y}\right) < \text{Var}(\hat{\lambda})$ |
| $n > 5$ therefore $n = 6$ | A1 | |
The image you've shared appears to be a **back/colophon page** of an Edexcel publication — it contains only publisher information (address, contact details, order code, and registration information), with no mark scheme content whatsoever.
There are **no questions, answers, mark allocations, or marking guidance** on this page to extract.
If you'd like me to extract mark scheme content, please share the **actual mark scheme pages** containing the questions and their corresponding answers/marks.6. Faults occur in a roll of material at a rate of $\lambda$ per $\mathrm { m } ^ { 2 }$. To estimate $\lambda$, three pieces of material of sizes $3 \mathrm {~m} ^ { 2 } , 7 \mathrm {~m} ^ { 2 }$ and $10 \mathrm {~m} ^ { 2 }$ are selected and the number of faults $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$ respectively are recorded.
The estimator $\hat { \lambda }$, where
$$\hat { \lambda } = k \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \right)$$
is an unbiased estimator of $\lambda$.
\begin{enumerate}[label=(\alph*)]
\item Write down the distributions of $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$ and find the value of $k$.
\item Find $\operatorname { Var } ( \hat { \lambda } )$.
A random sample of $n$ pieces of this material, each of size $4 \mathrm {~m} ^ { 2 }$, was taken. The number of faults on each piece, $Y$, was recorded.
\item Show that $\frac { 1 } { 4 } \bar { Y }$ is an unbiased estimator of $\lambda$.
\item Find $\operatorname { Var } \left( \frac { 1 } { 4 } \bar { Y } \right)$.
\item Find the minimum value of $n$ for which $\frac { 1 } { 4 } \bar { Y }$ becomes a better estimator of $\lambda$ than $\hat { \lambda }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2010 Q6 [14]}}