| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| 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 only direct application of the formula for parts (i)(A) and (i)(B), followed by a simple multiplication for the expected value in part (ii). The probability p=0.05 is small and n=8 is manageable, making calculations routine. No problem-solving insight or complex manipulation is needed—just standard textbook application of binomial probability and expectation. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 0) = 0.95^8 = 0.6634\) | M1, A1 | 0.663 or better; M1 \(0.95^8\); A1cao; Or B2 (tables) |
| Answer | Marks | Guidance |
|---|---|---|
| Part B: \(P(X = 1) = \binom{8}{1} \times 0.05 \times 0.95^7 = 0.2793\) | M1 | M1 for \(P(X = 1)\)(allow 0.28 or better) |
| \(P(X > 1) = 1 - (0.6634 + 0.2793) = 0.0573\) | M1, A1 | M1 for 1 − P(X ≤ 1) must have both probabilities; A1cao (0.0572 − 0.0573) |
| Or using tables: \(P(X > 1) = 1 - 0.9428 = 0.0572\) | M1, A1 | M1 for \(P(X \leq 1)\) 0.9428; M1 for 1 − P(X ≤ 1); A1cao (must end in ...2) |
| Answer | Marks | Guidance |
|---|---|---|
| Expected number of days \(= 250 \times 0.0572 = 14.3\) awrt | M1, A1 | M1 for \(250 \times \text{prob}(B)\); A1 FT but no rounding at end |
## (i)
**Part A:** $X \sim B(8, 0.05)$
$P(X = 0) = 0.95^8 = 0.6634$ | M1, A1 | 0.663 or better; M1 $0.95^8$; A1cao; Or B2 (tables)
Or using tables: $P(X = 0) = 0.6634$
**Part B:** $P(X = 1) = \binom{8}{1} \times 0.05 \times 0.95^7 = 0.2793$ | M1 | M1 for $P(X = 1)$(allow 0.28 or better)
$P(X > 1) = 1 - (0.6634 + 0.2793) = 0.0573$ | M1, A1 | M1 for 1 − P(X ≤ 1) must have both probabilities; A1cao (0.0572 − 0.0573)
Or using tables: $P(X > 1) = 1 - 0.9428 = 0.0572$ | M1, A1 | M1 for $P(X \leq 1)$ 0.9428; M1 for 1 − P(X ≤ 1); A1cao (must end in ...2) | 3 marks total
## (ii)
Expected number of days $= 250 \times 0.0572 = 14.3$ awrt | M1, A1 | M1 for $250 \times \text{prob}(B)$; A1 FT but no rounding at end | 2 marks total
---4 A small business has 8 workers. On a given day, the probability that any particular worker is off sick is 0.05 , independently of the other workers.
\begin{enumerate}[label=(\roman*)]
\item A day is selected at random. Find the probability that\\
(A) no workers are off sick,\\
(B) more than one worker is off sick.
\item There are 250 working days in a year. Find the expected number of days in the year on which more than one worker is off sick.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2008 Q4 [7]}}