| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Single batch expected count |
| Difficulty | Moderate -0.8 This is a straightforward binomial distribution question requiring direct application of the formula for P(X=1) and P(X>1)=1-P(X≤1), followed by a simple expected value calculation (multiply probability by 240). All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure and calculation requirements. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim B(50, 0.03)\) | ||
| \(P(X=1) = \binom{50}{1} \times 0.03 \times 0.97^{49} = 0.3372\) | M1 | \(0.03 \times 0.97^{49}\) or 0.0067(4)... |
| M1 | \(\binom{50}{1} \times pq^{49}\) \((p+q=1)\) | |
| A1 CAO | awfw 0.337 to 0.3372 or 0.34(2s.f.) or 0.34(2d.p.) but not just 0.34 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=0) = 0.97^{50} = 0.2181\) | B1 | awfw 0.218 to 0.2181 |
| \(P(X>1) = 1 - 0.2181 - 0.3372 = 0.4447\) | M1 | For \(1 - (\text{their } p(X=0) + \text{their } p(X=1))\); must have both probabilities |
| A1 CAO | awfw 0.4447 to 0.445 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Expected number \(= np = 240 \times 0.3372 = 80.88 - 80.93 = (81)\) | M1 | For \(240 \times \text{prob (A)}\) |
| Condone \(240 \times 0.34 = 81.6 = (82)\) but for M1 A1 f.t. | A1 FT | |
| TOTAL | 8 |
## Question 4:
### Part (i)(A):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim B(50, 0.03)$ | | |
| $P(X=1) = \binom{50}{1} \times 0.03 \times 0.97^{49} = 0.3372$ | M1 | $0.03 \times 0.97^{49}$ or 0.0067(4)... |
| | M1 | $\binom{50}{1} \times pq^{49}$ $(p+q=1)$ |
| | A1 CAO | awfw 0.337 to 0.3372 or 0.34(2s.f.) or 0.34(2d.p.) but not just 0.34 |
| **Total** | **3** | |
### Part (i)(B):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=0) = 0.97^{50} = 0.2181$ | B1 | awfw 0.218 to 0.2181 |
| $P(X>1) = 1 - 0.2181 - 0.3372 = 0.4447$ | M1 | For $1 - (\text{their } p(X=0) + \text{their } p(X=1))$; must have both probabilities |
| | A1 CAO | awfw 0.4447 to 0.445 |
| **Total** | **3** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Expected number $= np = 240 \times 0.3372 = 80.88 - 80.93 = (81)$ | M1 | For $240 \times \text{prob (A)}$ |
| Condone $240 \times 0.34 = 81.6 = (82)$ but for M1 A1 f.t. | A1 FT | |
| **TOTAL** | **8** | |
---4 A pottery manufacturer makes teapots in batches of 50. On average 3\% of teapots are faulty.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that in a batch of 50 there is\\
(A) exactly one faulty teapot,\\
(B) more than one faulty teapot.
\item The manufacturer produces 240 batches of 50 teapots during one month. Find the expected number of batches which contain exactly one faulty teapot.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2009 Q4 [8]}}