| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| 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 binomial with clear parameters (n=10500, p=0.0002 gives λ=2.1). Part (i) requires stating the well-known condition n large, p small; parts (ii) and (iii) involve routine Poisson probability calculations using tables. No problem-solving or novel insight required—purely procedural application of a standard approximation. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model |
| Answer | Marks |
|---|---|
| (i) Poisson | B1 |
| (Actually binomial with) | B1 |
| \(n > 50\) | B1 |
| and \(np\) (or \(\lambda\)) (= 2.1) which is \(< 5\) |
| Answer | Marks |
|---|---|
| (ii) \(\lambda = 2.1\) | B1 |
| \(e^{-2.1}\left(1 + 2.1 + \frac{2.1^2}{2} + \frac{2.1^3}{3!}\right) = 0.839\) (3 sf) | M1 A1 |
| Answer | Marks |
|---|---|
| (iii) \(P(X > 1) = 1 - e^{-2.1}\) (= 0.87754) | M1 |
| \(P(X = 1, 2, 3) = e^{-2.1}\left(2.1 + \frac{2.1^2}{2} + \frac{2.1^3}{3!}\right) = 0.71619\) | M1 |
| \(\frac{P(X = 1, 2, 3)}{P(X > 1)} = \frac{0.71619}{0.87754} = 0.816\) (3 sf) | M1 A1 |
(i) Poisson | B1
(Actually binomial with) | B1
$n > 50$ | B1
and $np$ (or $\lambda$) (= 2.1) which is $< 5$ |
Allow without "binomial"
Accept $n$ large
Accept $p$ small ($p < 0.1$)
**Total: 3**
(ii) $\lambda = 2.1$ | B1
$e^{-2.1}\left(1 + 2.1 + \frac{2.1^2}{2} + \frac{2.1^3}{3!}\right) = 0.839$ (3 sf) | M1 A1
Attempt $P(0, 1, 2, 3)$ any $\lambda$ allow 1 end error
SR1 Ft Normal $N(2.1, 2.1)$ B1 standardising M1; 0.833 A1
SR2 Ft Binomial $B(10500, 0.0002)$ B1 calculating binomial prob $P(0, 1, 2, 3)$ M1 = 0.8386 A1
**Total: 3**
(iii) $P(X > 1) = 1 - e^{-2.1}$ (= 0.87754) | M1
$P(X = 1, 2, 3) = e^{-2.1}\left(2.1 + \frac{2.1^2}{2} + \frac{2.1^3}{3!}\right) = 0.71619$ | M1
$\frac{P(X = 1, 2, 3)}{P(X > 1)} = \frac{0.71619}{0.87754} = 0.816$ (3 sf) | M1 A1
Any $\lambda$
Or '0.839' $- e^{-2.1}$
Any $\lambda$
Allow any attempted $P(X = 1, 2, 3)$ / $P(X > 1)$
SR1 Ft Normal $P(> 0.5) = 0.86523$ M1; $P(1, 2, 3) = 0.698$ M1; $0.698 /7 In a certain lottery, 10500 tickets have been sold altogether and each ticket has a probability of 0.0002 of winning a prize. The random variable $X$ denotes the number of prize-winning tickets that have been sold.\\
(i) State, with a justification, an approximating distribution for $X$.\\
(ii) Use your approximating distribution to find $\mathrm { P } ( X < 4 )$.\\
(iii) Use your approximating distribution to find the conditional probability that $X < 4$, given that $X \geqslant 1$.
\hfill \mbox{\textit{CAIE S2 2015 Q7 [10]}}