| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Multiple binomial probability calculations |
| Difficulty | Moderate -0.3 This is a straightforward application of binomial probability using tables, requiring multiple lookups and basic probability calculations. Part (ii) involves recognizing a binomial-of-binomial structure, but the steps are mechanical once identified. Slightly easier than average due to table usage and standard multi-part structure. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - 0.7899 = 0.210(1)\) | M1 | For complement of relevant tabular value |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.9209 - 0.7899 = 0.131\) | M1 | For subtracting relevant tabular values |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.790^3 + 5 \times 0.790^4 \times 0.210 + 10 \times 0.790^3 \times 0.210^2\) | M1 | For recognition of \(B(5, 0.210)\) |
| M1 | For identification of correct three cases | |
| A1√ | For correct expression for the required prob | |
| A1 | For correct answer | |
| \(= 0.934\) |
| Answer | Marks | Guidance |
|---|---|---|
| Expectation is \(5 \times 0.210 = 1.05\) | M1 | For relevant use of \(np\) |
| A1 | For correct answer |
**Part (i)(a)**
$1 - 0.7899 = 0.210(1)$ | M1 | For complement of relevant tabular value
| A1 | For correct answer
**Part (i)(b)**
$0.9209 - 0.7899 = 0.131$ | M1 | For subtracting relevant tabular values
| A1 | For correct answer
**Part (ii)(a)**
$0.790^3 + 5 \times 0.790^4 \times 0.210 + 10 \times 0.790^3 \times 0.210^2$ | M1 | For recognition of $B(5, 0.210)$
| M1 | For identification of correct three cases
| A1√ | For correct expression for the required prob
| A1 | For correct answer
$= 0.934$ |
**Part (ii)(b)**
Expectation is $5 \times 0.210 = 1.05$ | M1 | For relevant use of $np$
| A1 | For correct answer
---7 Items from a production line are examined for any defects. The probability that any item will be found to be defective is 0.15 , independently of all other items.\\
(i) A batch of 16 items is inspected. Using tables of cumulative binomial probabilities, or otherwise, find the probability that
\begin{enumerate}[label=(\alph*)]
\item at least 4 items in the batch are defective,
\item exactly 4 items in the batch are defective.\\
(ii) Five batches, each containing 16 items, are taken.\\
(a) Find the probability that at most 2 of these 5 batches contain at least 4 defective items.\\
(b) Find the expected number of batches that contain at least 4 defective items.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 Q7 [10]}}