| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Percentage error in approximation |
| Difficulty | Standard +0.3 This question tests standard application of Poisson approximation to binomial with straightforward percentage error calculation in part (a), and a routine one-tailed hypothesis test in part (b). Both parts require only direct application of learned procedures with no novel problem-solving, making it slightly easier than average for Further Maths Statistics. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 4.5\) | B1 | |
| \(e^{-4.5}\) \((= 0.011109)\) | M1 | Alone allow any \(\lambda\) |
| \(\left(\frac{99}{100}\right)^{450}\) \((= 0.010860)\) | M1 | |
| \(\frac{(\text{'0.011109'} - \text{'0.010860'})}{\text{'0.010860'}} \times 100\) | ||
| \(= 2.29\%\) (3 sf) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: P(6) = \frac{1}{6}\) or \(p = \frac{1}{6}\); \(H_1: P(6) < \frac{1}{6}\) or \(p < \frac{1}{6}\) | B1 | Both needed |
| \(\left(\frac{5}{6}\right)^{25} + 25\left(\frac{5}{6}\right)^{24} \times \frac{1}{6} + {}^{25}C_2\left(\frac{5}{6}\right)^{23} \times \left(\frac{1}{6}\right)^2\) | M1 | Allow one error (extra/missing term/incorrect term); CR method: attempt at least \(P(0)\) and \(P(1)\) \((0.010\ldots\) and \(0.06\ldots < 0.1)\) |
| \(= 0.189\) (3 sf) | A1 | CR is 0,1 and must see 0.189 for A1 |
| comp \(0.1\) | M1 | Valid comparison '0.189' with 0.1 oe; valid comparison of 2 with CR |
| No reason to believe die biased | A1 [5] | Correct conclusion \(\checkmark\) their 0.189; no contradictions |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 4.5$ | B1 | |
| $e^{-4.5}$ $(= 0.011109)$ | M1 | Alone allow any $\lambda$ |
| $\left(\frac{99}{100}\right)^{450}$ $(= 0.010860)$ | M1 | |
| $\frac{(\text{'0.011109'} - \text{'0.010860'})}{\text{'0.010860'}} \times 100$ | | |
| $= 2.29\%$ (3 sf) | A1 [4] | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: P(6) = \frac{1}{6}$ or $p = \frac{1}{6}$; $H_1: P(6) < \frac{1}{6}$ or $p < \frac{1}{6}$ | B1 | Both needed |
| $\left(\frac{5}{6}\right)^{25} + 25\left(\frac{5}{6}\right)^{24} \times \frac{1}{6} + {}^{25}C_2\left(\frac{5}{6}\right)^{23} \times \left(\frac{1}{6}\right)^2$ | M1 | Allow one error (extra/missing term/incorrect term); CR method: attempt at least $P(0)$ and $P(1)$ $(0.010\ldots$ and $0.06\ldots < 0.1)$ |
| $= 0.189$ (3 sf) | A1 | CR is 0,1 and must see 0.189 for A1 |
| comp $0.1$ | M1 | Valid comparison '0.189' with 0.1 oe; valid comparison of 2 with CR |
| No reason to believe die biased | A1 [5] | Correct conclusion $\checkmark$ their 0.189; no contradictions |
---5
\begin{enumerate}[label=(\alph*)]
\item Narika has a die which is known to be biased so that the probability of throwing a 6 on any throw is $\frac { 1 } { 100 }$. She uses an approximating distribution to calculate the probability of obtaining no 6s in 450 throws. Find the percentage error in using the approximating distribution for this calculation.
\item Johan claims that a certain six-sided die is biased so that it shows a 6 less often than it would if the die were fair. In order to test this claim, the die is thrown 25 times and it shows a 6 on only 2 throws. Test at the $10 \%$ significance level whether Johan's claim is justified.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2015 Q5 [9]}}