| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Multiple independent binomial calculations |
| Difficulty | Moderate -0.3 This question involves straightforward binomial probability calculations with clearly stated probabilities and sample sizes. Part (i) requires direct application of binomial probability formulas for small n, while part (ii) is a standard normal approximation to binomial with large n=240. All steps are routine S1 techniques with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0,1,2) = {^6C_0}\, 0.3^0\, 0.7^6 + {^6C_1}\, 0.3^1\, 0.7^5 + {^6C_2}\, 0.3^2\, 0.7^4\) | M1 | Binomial term of form \(^6C_x p^x(1-p)^{6-x}\), \(0 < p < 1\), any \(p\), \(x \neq 6, 0\) |
| \(0.1176\ldots + 0.3025\ldots + 0.3241\ldots\) | A1 | Correct unsimplified answer |
| \(0.744\) | A1 | Correct final answer |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{support neither choir}) = 1-(0.3+0.45) = 0.25\) | M1 | \(0.25^n\) seen alone, \(1 < n \leqslant 6\) |
| \(P(\text{6 support neither choir}) = 0.25^6 = 0.000244\) or \(\frac{1}{4096}\) | A1 | Correct final answer |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 240 \times 0.25 = 60\); Variance \(= 240 \times 0.25 \times 0.75 = 45\) | B1FT | Correct unsimplified \(240p\) and \(240pq\) where \(p = \textit{their}\) P(support neither choir) or 0.25 |
| \(P(X < 50) = P\!\left(Z < \frac{49.5-60}{\sqrt{45}}\right) = P(Z < -1.565)\) | M1 | Substituting *their* \(\mu\) and \(\sigma\) (condone \(\sigma^2\)) into the \(\pm\)Standardisation Formula with a numerical value for '\(49.5\)' |
| M1 | Using continuity correction 49.5 or 50.5 within a standardisation expression | |
| \(1 - 0.9412\) | M1 | Appropriate area \(\Phi\) from standardisation formula \(P(z<\ldots)\) in final solution, (\(< 0.5\) if \(z\) is \(-\)ve, \(> 0.5\) if \(z\) is \(+\)ve) |
| \(0.0588\) | A1 | Correct final answer |
| 5 |
## Question 7(i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0,1,2) = {^6C_0}\, 0.3^0\, 0.7^6 + {^6C_1}\, 0.3^1\, 0.7^5 + {^6C_2}\, 0.3^2\, 0.7^4$ | M1 | Binomial term of form $^6C_x p^x(1-p)^{6-x}$, $0 < p < 1$, any $p$, $x \neq 6, 0$ |
| $0.1176\ldots + 0.3025\ldots + 0.3241\ldots$ | A1 | Correct unsimplified answer |
| $0.744$ | A1 | Correct final answer |
| | **3** | |
---
## Question 7(i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{support neither choir}) = 1-(0.3+0.45) = 0.25$ | M1 | $0.25^n$ seen alone, $1 < n \leqslant 6$ |
| $P(\text{6 support neither choir}) = 0.25^6 = 0.000244$ or $\frac{1}{4096}$ | A1 | Correct final answer |
| | **2** | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 240 \times 0.25 = 60$; Variance $= 240 \times 0.25 \times 0.75 = 45$ | B1FT | Correct unsimplified $240p$ and $240pq$ where $p = \textit{their}$ P(support neither choir) or 0.25 |
| $P(X < 50) = P\!\left(Z < \frac{49.5-60}{\sqrt{45}}\right) = P(Z < -1.565)$ | M1 | Substituting *their* $\mu$ and $\sigma$ (condone $\sigma^2$) into the $\pm$Standardisation Formula with a numerical value for '$49.5$' |
| | M1 | Using continuity correction 49.5 or 50.5 within a standardisation expression |
| $1 - 0.9412$ | M1 | Appropriate area $\Phi$ from standardisation formula $P(z<\ldots)$ in final solution, ($< 0.5$ if $z$ is $-$ve, $> 0.5$ if $z$ is $+$ve) |
| $0.0588$ | A1 | Correct final answer |
| | **5** | |7 A competition is taking place between two choirs, the Notes and the Classics. There is a large audience for the competition.
\begin{itemize}
\item $30 \%$ of the audience are Notes supporters.
\item $45 \%$ of the audience are Classics supporters.
\item The rest of the audience are not supporters of either of these choirs.
\item No one in the audience supports both of these choirs.
\begin{enumerate}[label=(\roman*)]
\item A random sample of 6 people is chosen from the audience.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that no more than 2 of the 6 people are Notes supporters.
\item Find the probability that none of the 6 people support either of these choirs.
\item A random sample of 240 people is chosen from the audience. Use a suitable approximation to find the probability that fewer than 50 do not support either of the choirs.\\
\end{itemize}
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2019 Q7 [10]}}