| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | State Poisson approximation with justification |
| Difficulty | Moderate -0.8 This is a straightforward application of the standard Poisson approximation to the binomial distribution. Part (i) requires stating B(520, 0.008) and Po(4.16) with the standard justification (n large, p small, np moderate). Part (ii) involves routine Poisson probability calculations using tables or formulas. The question tests recall of conditions and basic computation rather than problem-solving or insight. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(B(520, 0.008)\); \(Po(4.16)\) | B1, B1B1 | Po: B1, \(\lambda = 4.16\): B1 |
| \(n = 500\) which is large; \(np = 4.16\) which is \(< 5\) or \(p\) small \(< 0.1\) | B1 [4] | Both needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - e^{-4.16}\!\left(1 + 4.16 + \dfrac{4.16^2}{2} + \dfrac{4.16^3}{3!}\right) = 0.597\) (3 sf) | M1, A1 [2] | \(1 - P(0,1,2,3)\) any \(\lambda\); allow one end error |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-4.16} \times \dfrac{4.16^n}{n!} > e^{-4.16} \times \dfrac{4.16^{n+1}}{(n+1)!}\) | M1 | any \(\lambda\) |
| \(1 > \dfrac{4.16}{n+1}\); \(n > 3.16\) | A1 | or equiv equation without \(e\) and without factorials |
| Smallest \(n\) is 4 | A1 [3] | Calculation of \(P(0), P(1),\ldots P(5)\) scores M1 for at least 3 attempted, A1 all correct, A1 for \(n = 4\) |
## Question 5:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $B(520, 0.008)$; $Po(4.16)$ | B1, B1B1 | Po: B1, $\lambda = 4.16$: B1 |
| $n = 500$ which is large; $np = 4.16$ which is $< 5$ or $p$ small $< 0.1$ | B1 [4] | Both needed |
### Part (ii)(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - e^{-4.16}\!\left(1 + 4.16 + \dfrac{4.16^2}{2} + \dfrac{4.16^3}{3!}\right) = 0.597$ (3 sf) | M1, A1 [2] | $1 - P(0,1,2,3)$ any $\lambda$; allow one end error |
### Part (ii)(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-4.16} \times \dfrac{4.16^n}{n!} > e^{-4.16} \times \dfrac{4.16^{n+1}}{(n+1)!}$ | M1 | any $\lambda$ |
| $1 > \dfrac{4.16}{n+1}$; $n > 3.16$ | A1 | or equiv equation without $e$ and without factorials |
| Smallest $n$ is 4 | A1 [3] | Calculation of $P(0), P(1),\ldots P(5)$ scores M1 for at least 3 attempted, A1 all correct, A1 for $n = 4$ |
---5 The probability that a new car of a certain type has faulty brakes is 0.008 . A random sample of 520 new cars of this type is chosen, and the number, $X$, having faulty brakes is noted.\\
(i) Describe fully the distribution of $X$ and describe also a suitable approximating distribution. Justify this approximating distribution.\\
(ii) Use your approximating distribution to find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( X > 3 )$,
\item the smallest value of $n$ such that $\mathrm { P } ( X = n ) > \mathrm { P } ( X = n + 1 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2013 Q5 [9]}}