| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Challenging +1.2 This question requires understanding of Poisson distribution properties including parameter estimation from probability ratios, scaling for different lengths, and normal approximation for large parameters. Part (i) involves solving an equation from P(X=2)=3P(X=4), then scaling λ for 3.5m. Part (ii)(a) requires using P(X≥1)=1-P(X=0) and logarithms. Part (ii)(b) needs normal approximation with continuity correction. While multi-step and covering several techniques, these are standard S2 applications without requiring novel insight—moderately above average difficulty for A-level but routine for Further Maths Statistics. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{e^{-\lambda}\lambda^2}{2!} = 3\frac{e^{-\lambda}\lambda^4}{4!}\) | M1 | Poisson equation involving \(\lambda\). |
| \(\lambda = 2\) | A1 | Correct mean |
| New \(\lambda = 7\) | B1 ft | New mean ft 3.5 × previous one |
| \(P(X > 3) = 1 - e^{-7}\left[1 + 7 + \frac{7^2}{2!} + \frac{7^3}{3!}\right] = 0.918\) | M1 | Poisson probs with their mean (at least 3 probs) and 1- |
| A1 | 5 | Correct answer |
| (ii)(a) \(\lambda = 1.3k\) | B1 | Correct new mean |
| \(P(X > 0) = 1 - e^{-1.3k} = 0.96\) | M1 | Equation with k or λ in involving \(1 - P(0) = 0.96\) |
| \(0.04 = e^{-1.3k}\) | A1 | correct equation |
| \(k = 2.48\) | A1 | 4 |
| (b) \(X \sim N(1300, 1300)\) | B1 | correct mean and variance |
| \(P(X > 1250) = P\left(z > \frac{1250.5 - 1300}{\sqrt{1300}}\right)\) | M1 | standardising must have sq rt with or without cc |
| \(= P(z > -1.373) = 0.915\) | A1 | 3 |
**(i)** $\frac{e^{-\lambda}\lambda^2}{2!} = 3\frac{e^{-\lambda}\lambda^4}{4!}$ | M1 | Poisson equation involving $\lambda$.
$\lambda = 2$ | A1 | Correct mean
New $\lambda = 7$ | B1 ft | New mean ft 3.5 × previous one
$P(X > 3) = 1 - e^{-7}\left[1 + 7 + \frac{7^2}{2!} + \frac{7^3}{3!}\right] = 0.918$ | M1 | Poisson probs with their mean (at least 3 probs) and 1-
| A1 | 5 | Correct answer
**(ii)(a)** $\lambda = 1.3k$ | B1 | Correct new mean
$P(X > 0) = 1 - e^{-1.3k} = 0.96$ | M1 | Equation with k or λ in involving $1 - P(0) = 0.96$
$0.04 = e^{-1.3k}$ | A1 | correct equation
$k = 2.48$ | A1 | 4 | correct answer
**(b)** $X \sim N(1300, 1300)$ | B1 | correct mean and variance
$P(X > 1250) = P\left(z > \frac{1250.5 - 1300}{\sqrt{1300}}\right)$ | M1 | standardising must have sq rt with or without cc
$= P(z > -1.373) = 0.915$ | A1 | 3 | correct answer6 The random variable $X$ denotes the number of worms on a one metre length of a country path after heavy rain. It is given that $X$ has a Poisson distribution.\\
(i) For one particular path, the probability that $X = 2$ is three times the probability that $X = 4$. Find the probability that there are more than 3 worms on a 3.5 metre length of this path.\\
(ii) For another path the mean of $X$ is 1.3.
\begin{enumerate}[label=(\alph*)]
\item On this path the probability that there is at least 1 worm on a length of $k$ metres is 0.96 . Find $k$.
\item Find the probability that there are more than 1250 worms on a one kilometre length of this path.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2007 Q6 [12]}}