| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths Statistics question requiring knowledge of Poisson distribution properties, including the sum of independent Poisson variables and solving for parameters. Part (c) requires algebraic manipulation using the given probability equation and recognizing that X+Y ~ Poisson(λ_X + λ_Y). While it involves several steps and conceptual understanding beyond basic A-level, the techniques are standard for Further Statistics and don't require novel insight—making it moderately above average difficulty. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.168(1)\) | B1 [1] | Awrt 0.168. Allow \(\frac{27}{8}e^{-3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2 < X < 8) = 0.5649\ldots\) | M1 | M1A0 for 0.341 or 0.573 or 0.647 or 0.789 |
| A1 | Awrt 0.565, can be implied | |
| \(10 \times 0.565 = 5.65\) | M1 [3] | \(10\times\) *their* 0.565, allow rounded to nearest integer only if unrounded value seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X + Y \sim \text{Po}(3 + \lambda)\) | M1 | Any appropriate symbol, stated or implied, allow sign errors |
| \(e^{-3-\lambda}\frac{(3+\lambda)^4}{4!} = \frac{27}{8}e^{-3}\frac{3^2}{2!}e^{-\lambda}\frac{\lambda^2}{2!}\) DR | M1 | One correct use of Poisson formula anywhere |
| A1 | Correct equation, allow \(\frac{27}{8} \times 0.224 \times e^{-\lambda}\frac{\lambda^2}{2!}\) or \(0.378\lambda^2 e^{-\lambda}\) oe | |
| \(4(3+\lambda)^4 = 729\lambda^2\) | M1 | Cancel \(e\) terms and simplify factorials, e.g. \((3+\lambda)^4 = 182.25\lambda^2\) (correct treatment of \(e^{-3}\) can be implied by correct solutions) |
| \(2(3+\lambda)^2 = 27\lambda\) *or* \(4\lambda^4 + 48\lambda^3 - 513\lambda^2 + 432\lambda + 324 = 0\) | M1 | Take square root, allow ignoring negatives, find at least one solution to quadratic *or* multiply out to obtain quartic *and solve 5-term equation* |
| No need to consider other sign as \(\lambda > 0\) | B1 | Explicit reason for rejection of \(-\) sign or of negative solutions to quartic (other solutions are \(-\frac{3}{4}(13 \pm 3\sqrt{17}) = -0.473\) or \(-19.027\)) |
| \(2\lambda^2 - 15\lambda + 18 = 0 \Rightarrow \lambda = 6\) or \(\frac{3}{2}\) | B1 [7] | Both solutions correct, www, no others. SC: \(4(3+\lambda)^4 = 729\lambda^2\) seen and both solutions correct but no working: (M1M1A1) M0B0B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| In using \(X + Y \sim \text{Po}(3 + \lambda)\) | B1 [1] | Allow \(\text{Po}(X+Y) = \text{Po}(X) + \text{Po}(Y)\), or \(P(X+Y=4) = P(X=4) \times P(Y=4)\), but *not* \(E(X+Y) = E(X) + E(Y)\), *nor* \(P(X+Y=4) = P(X=4) \times P(Y=4)\), *nor* "when adding parameters". "When doing \(P(X+Y)\) assumed probability of one not affecting other": B1. Any wrong statement: B0 |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.168(1)$ | **B1 [1]** | Awrt 0.168. Allow $\frac{27}{8}e^{-3}$ |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2 < X < 8) = 0.5649\ldots$ | **M1** | M1A0 for 0.341 or 0.573 or 0.647 or 0.789 |
| | **A1** | Awrt 0.565, can be implied |
| $10 \times 0.565 = 5.65$ | **M1 [3]** | $10\times$ *their* 0.565, allow rounded to nearest integer only if unrounded value seen |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X + Y \sim \text{Po}(3 + \lambda)$ | **M1** | Any appropriate symbol, stated or implied, allow sign errors |
| $e^{-3-\lambda}\frac{(3+\lambda)^4}{4!} = \frac{27}{8}e^{-3}\frac{3^2}{2!}e^{-\lambda}\frac{\lambda^2}{2!}$ **DR** | **M1** | One correct use of Poisson formula anywhere |
| | **A1** | Correct equation, allow $\frac{27}{8} \times 0.224 \times e^{-\lambda}\frac{\lambda^2}{2!}$ or $0.378\lambda^2 e^{-\lambda}$ oe |
| $4(3+\lambda)^4 = 729\lambda^2$ | **M1** | Cancel $e$ terms and simplify factorials, e.g. $(3+\lambda)^4 = 182.25\lambda^2$ (correct treatment of $e^{-3}$ can be implied by correct solutions) |
| $2(3+\lambda)^2 = 27\lambda$ *or* $4\lambda^4 + 48\lambda^3 - 513\lambda^2 + 432\lambda + 324 = 0$ | **M1** | Take square root, allow ignoring negatives, find at least one solution to quadratic *or* multiply out to obtain quartic *and solve 5-term equation* |
| No need to consider other sign as $\lambda > 0$ | **B1** | Explicit reason for rejection of $-$ sign or of negative solutions to quartic (other solutions are $-\frac{3}{4}(13 \pm 3\sqrt{17}) = -0.473$ or $-19.027$) |
| $2\lambda^2 - 15\lambda + 18 = 0 \Rightarrow \lambda = 6$ or $\frac{3}{2}$ | **B1 [7]** | Both solutions correct, www, no others. SC: $4(3+\lambda)^4 = 729\lambda^2$ seen and both solutions correct but no working: (M1M1A1) M0B0B1 |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| In using $X + Y \sim \text{Po}(3 + \lambda)$ | **B1 [1]** | Allow $\text{Po}(X+Y) = \text{Po}(X) + \text{Po}(Y)$, or $P(X+Y=4) = P(X=4) \times P(Y=4)$, but *not* $E(X+Y) = E(X) + E(Y)$, *nor* $P(X+Y=4) = P(X=4) \times P(Y=4)$, *nor* "when adding parameters". "When doing $P(X+Y)$ assumed probability of one not affecting other": B1. Any wrong statement: B0 |
---\begin{enumerate}[label=(\alph*)]
\item Find the probability that 4 telephone calls are received in a randomly chosen one-minute period.
\item A sample of 10 independent observations of $X$ is obtained.
Find the expected number of these 10 observations that are in the interval $2 < X < 8$.
It is also known that\\
$P ( X + Y = 4 ) = \frac { 27 } { 8 } P ( X = 2 ) \times P ( Y = 2 )$.
\item Determine the possible values of $\mathrm { E } ( Y )$.
\item Explain where in your solution to part (c) you have used the assumption that telephone calls and e-mails are received independently of one another.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2024 Q4 [12]}}