| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Identify distribution and parameters |
| Difficulty | Moderate -0.8 This is a straightforward binomial distribution question requiring only identification of parameters (n=12, p=0.1), standard assumptions, and direct application of binomial probability formulas. Part (iii) involves recognizing a second binomial distribution but uses routine complement and summation techniques. No problem-solving insight or complex manipulation required—purely procedural S1 content. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Binomial | B1 | |
| \(n = 12,\ p = 0.1\) | B1 | \(B(12, 0.1)\): B2 |
| Plates (or seconds) independent | B1 | NOT: batches independent |
| Prob of fault same for each plate | B1 4 | Comments must be in context; ignore incorrect or irrelevant |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.9744 - 0.8891\) or \(^{12}C_3 \times 0.9^9 \times 0.1^3 = 0.0852\) or \(0.0853\) (3 sfs) | M1, A1 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - 0.2824\) or \(1 - 0.9^{12} = 0.718\) (3 sfs) | M1, A1 2 | allow \(1 - 0.6590\) or \(1 - 0.9^{11}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| "0.718" and \(1 -\) "0.718" used | B1 | ft (b) for B1M1M1 |
| \((1-0.718)^4 + 4(1-0.718)^3 \times 0.718 + ^4C_2(1-0.718)^2 \times 0.718^2\) | M2 | M1 for any one term correct (eg opposite tail or no coefficients); \(1 - P(3\) or \(4)\) follow similar scheme M2 or M1; \(1 -\) correct working \((= 0.623)\): B1M2 |
| \(= 0.317\) (3 sfs) | A1 4 | cao |
## Question 7:
**Part (i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Binomial | B1 | |
| $n = 12,\ p = 0.1$ | B1 | $B(12, 0.1)$: B2 |
| Plates (or seconds) independent | B1 | NOT: batches independent |
| Prob of fault same for each plate | B1 4 | Comments must be in context; ignore incorrect or irrelevant |
**Part (ii)(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.9744 - 0.8891$ or $^{12}C_3 \times 0.9^9 \times 0.1^3 = 0.0852$ or $0.0853$ (3 sfs) | M1, A1 2 | |
**Part (ii)(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - 0.2824$ or $1 - 0.9^{12} = 0.718$ (3 sfs) | M1, A1 2 | allow $1 - 0.6590$ or $1 - 0.9^{11}$ |
**Part (iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| "0.718" and $1 -$ "0.718" used | B1 | ft (b) for B1M1M1 |
| $(1-0.718)^4 + 4(1-0.718)^3 \times 0.718 + ^4C_2(1-0.718)^2 \times 0.718^2$ | M2 | M1 for any one term correct (eg opposite tail or no coefficients); $1 - P(3$ or $4)$ follow similar scheme M2 or M1; $1 -$ correct working $(= 0.623)$: B1M2 |
| $= 0.317$ (3 sfs) | A1 4 | cao |
---7 At a factory that makes crockery the quality control department has found that $10 \%$ of plates have minor faults. These are classed as 'seconds'. Plates are stored in batches of 12. The number of seconds in a batch is denoted by $X$.\\
(i) State an appropriate distribution with which to model $X$. Give the value(s) of any parameter(s) and state any assumptions required for the model to be valid.
Assume now that your model is valid.\\
(ii) Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( X = 3 )$,
\item $\mathrm { P } ( X \geqslant 1 )$.\\
(iii) A random sample of 4 batches is selected. Find the probability that the number of these batches that contain at least 1 second is fewer than 3 .
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2009 Q7 [12]}}