| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Standard +0.3 This is a straightforward S2 Poisson question requiring standard techniques: writing Poisson probabilities, scaling parameters for different time periods, and using normal approximation with continuity correction. Part (c)(ii) involves solving λ from a z-score equation, which is routine algebra. All steps are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(Po(\lambda)\); \(e^{-\lambda}+\lambda e^{-\lambda}\) | M1, A1oe | Writing or using \(Po(\lambda)\); any correct equivalent expression |
| Answer | Marks | Guidance |
|---|---|---|
| \(Po(\frac{\lambda}{2})\); \(\frac{\lambda}{2}e^{-\lambda/2}\) | M1, A1oe | Writing or using \(Po(\lambda/2)\); any correct equivalent expression |
| Answer | Marks | Guidance |
|---|---|---|
| Mean is large, so a normal approximation is used. | B1 | Allow 'since mean \(10\lambda\) is large' |
| Answer | Marks | Guidance |
|---|---|---|
| \(Y\sim N(10\lambda, 10\lambda)\); \(P(Y<15)=P\!\left(Z<\frac{14.5-10\lambda}{\sqrt{10\lambda}}\right)\); \(\frac{14.5-10\lambda}{\sqrt{10\lambda}}=-2.1(0)\); \(10\lambda-2.10\sqrt{10\lambda}-14.5=0\to\sqrt{10\lambda}=\frac{2.10\pm\sqrt{2.10^2-4(-14.5)}}{2}=5\); or \(100\lambda^2-334.1\lambda+210.25=0\to\lambda=2.5\) or \(0.841\); \(\lambda=2.5\) | M1, M1, M1 M1 A1, dM1 A1, A1 | 1st M1: mean=variance \([=10\lambda]\); 2nd M1: continuity correction 14.5 or 15.5; 3rd M1: standardising with 14.5/15/15.5 and s.d.\(=\sqrt{\text{mean}}\), equating to \(z\)-value \( |
# Question 6:
## Part (a)
| $Po(\lambda)$; $e^{-\lambda}+\lambda e^{-\lambda}$ | M1, A1oe | Writing or using $Po(\lambda)$; any correct equivalent expression |
## Part (b)
| $Po(\frac{\lambda}{2})$; $\frac{\lambda}{2}e^{-\lambda/2}$ | M1, A1oe | Writing or using $Po(\lambda/2)$; any correct equivalent expression |
## Part (c)(i)
| Mean is large, so a normal approximation is used. | B1 | Allow 'since mean $10\lambda$ is large' |
## Part (c)(ii)
| $Y\sim N(10\lambda, 10\lambda)$; $P(Y<15)=P\!\left(Z<\frac{14.5-10\lambda}{\sqrt{10\lambda}}\right)$; $\frac{14.5-10\lambda}{\sqrt{10\lambda}}=-2.1(0)$; $10\lambda-2.10\sqrt{10\lambda}-14.5=0\to\sqrt{10\lambda}=\frac{2.10\pm\sqrt{2.10^2-4(-14.5)}}{2}=5$; or $100\lambda^2-334.1\lambda+210.25=0\to\lambda=2.5$ or $0.841$; $\lambda=2.5$ | M1, M1, M1 M1 A1, dM1 A1, A1 | 1st M1: mean=variance $[=10\lambda]$; 2nd M1: continuity correction 14.5 or 15.5; 3rd M1: standardising with 14.5/15/15.5 and s.d.$=\sqrt{\text{mean}}$, equating to $z$-value $|z|>1$; 1st A1: $\frac{14.5-\mu}{\sqrt{\mu}}=$ awrt $-2.10$; 4th dM1: solving their 3TQ; 2nd A1: $\sqrt{\mu}$ awrt 5.00 or $\mu$ awrt 25.0; 3rd A1: $\lambda=$ awrt **2.5(0)** (must reject other solutions) |\begin{enumerate}
\item According to an electric company, power failures occur randomly at a rate of $\lambda$ every 10 weeks, $1 < \lambda < 10$\\
(a) Write down an expression in terms of $\lambda$ for the probability that there are fewer than 2 power failures in a randomly selected 10 week period.\\
(b) Write down an expression in terms of $\lambda$ for the probability that there is exactly 1 power failure in a randomly selected 5 week period.
\end{enumerate}
Over a 100 week period, the probability, using a normal approximation, that fewer than 15 power failures occur is 0.0179 (to 3 significant figures).\\
(c) (i) Justify the use of a normal approximation.\\
(ii) Find the value of $\lambda$.
Show each stage of your working clearly.\\
\hfill \mbox{\textit{Edexcel S2 2016 Q6 [12]}}