| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Standard +0.3 Part (i) is straightforward application of Poisson addition property and cumulative probability. Part (ii) requires setting up an equation using Poisson probability formula and solving for n, which involves algebraic manipulation but is routine. Part (iii) applies normal approximation to Poisson, a standard technique. All parts are textbook-style applications with no novel insight required, making this slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = 4.8\) | B1 | |
| \(E^{-4}(1 + 4.8 + \frac{4.8^2}{2!} + \frac{4.8^3}{3!}) = 0.294\) (3 sfs) | M1 A1 [3] | P(R = 0, 1, 2 or 3), their \(\lambda\) allow one end error |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-\lambda} \times \frac{4!}{4!} = \frac{16}{3} e^{-\lambda} \times \frac{2}{1}\) or without \(e^{-\lambda}\) | M1 | \(\lambda = 1.6n\) seen or implied; B1; \(e^{-1.6n} \times \frac{(1.6n)^4}{4!} = \frac{16}{3} e^{-1.6n} \times \frac{(1.6n)^2}{2!}\) |
| \(\frac{\lambda^2}{12} = \frac{16}{3}\) or better | A1 | \(\frac{(1.6n)^2}{12} = \frac{16}{3}\) or better |
| \((\lambda = 8)\); \(\lambda = 1.6n\) seen or implied; \(n = '8' ÷ 1.6 = 5\) | B1 A1 [4] | \(n = 5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(B \sim N(64, 64)\); \(\frac{75.5−64}{\sqrt{64}} (= 1.4375)\) | B1 M1 | Allow with wrong or no cc. No sd/var mixes |
| \(1 − \Phi('1.4375') (= 1 − 0.9247) = 0.0753\) to \(0.0754\) | M1 A1 [4] | Finding correct area consistent with their working |
**(i)**
$\lambda = 4.8$ | B1 |
$E^{-4}(1 + 4.8 + \frac{4.8^2}{2!} + \frac{4.8^3}{3!}) = 0.294$ (3 sfs) | M1 A1 [3] | P(R = 0, 1, 2 or 3), their $\lambda$ allow one end error
**(ii)**
$e^{-\lambda} \times \frac{4!}{4!} = \frac{16}{3} e^{-\lambda} \times \frac{2}{1}$ or without $e^{-\lambda}$ | M1 | $\lambda = 1.6n$ seen or implied; B1; $e^{-1.6n} \times \frac{(1.6n)^4}{4!} = \frac{16}{3} e^{-1.6n} \times \frac{(1.6n)^2}{2!}$ | M1
$\frac{\lambda^2}{12} = \frac{16}{3}$ or better | A1 | $\frac{(1.6n)^2}{12} = \frac{16}{3}$ or better | A1
$(\lambda = 8)$; $\lambda = 1.6n$ seen or implied; $n = '8' ÷ 1.6 = 5$ | B1 A1 [4] | $n = 5$ | A1
**(iii)**
$B \sim N(64, 64)$; $\frac{75.5−64}{\sqrt{64}} (= 1.4375)$ | B1 M1 | Allow with wrong or no cc. No sd/var mixes
$1 − \Phi('1.4375') (= 1 − 0.9247) = 0.0753$ to $0.0754$ | M1 A1 [4] | Finding correct area consistent with their working7 A random variable $X$ has the distribution $\operatorname { Po } ( 1.6 )$.\\
(i) The random variable $R$ is the sum of three independent values of $X$. Find $\mathrm { P } ( R < 4 )$.\\
(ii) The random variable $S$ is the sum of $n$ independent values of $X$. It is given that
$$\mathrm { P } ( S = 4 ) = \frac { 16 } { 3 } \times \mathrm { P } ( S = 2 )$$
Find $n$.\\
(iii) The random variable $T$ is the sum of 40 independent values of $X$. Find $\mathrm { P } ( T > 75 )$.
\hfill \mbox{\textit{CAIE S2 2012 Q7 [11]}}