| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | General probability threshold |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question requiring standard probability calculations. Part (i) involves computing P(2 < X < 12) = 1 - P(X ≤ 2) - P(X = 12) with clearly given parameters. Part (ii) requires solving 1 - (0.87)^n > 0.95 using logarithms, which is a routine complementary probability problem. Both parts are textbook applications with no conceptual challenges beyond basic binomial setup. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(2 < X \leq 12) = 1 - P(0, 1, 2, 12)\) | M1 | Using binomial with \(_nC_{\text{something}}\) and powers summing to 12, \(\Sigma p = 1\) |
| \(= 1 - (0.35)^{12} - (0.65)(0.35)^{11}_nC_1 - (0.65)^2(0.35)^{10}_nC_2 - (0.65)^{12}(0.35)^{10}_nC_2 - (0.65)^{12}\) | A1 | Correct unsimplified answer |
| \(= 1 - 0.0065359 = 0.993\) | A1 [3] | Accept 0.994 from correct working only |
| (ii) \(1 - (0.87)^n > 0.95\) | M1 | Equality or inequality in (0.87 or 0.78 or 0.35), power \(n\) or \(n - 1\), 0.95 or 0.05 |
| \(0.05 > (0.87)^n\) | M1 | Attempt to solve an equation with a power in (can be implied) |
| \(n = 22\) | A1 [3] | Correct answer |
(i) $P(2 < X \leq 12) = 1 - P(0, 1, 2, 12)$ | M1 | Using binomial with $_nC_{\text{something}}$ and powers summing to 12, $\Sigma p = 1$
$= 1 - (0.35)^{12} - (0.65)(0.35)^{11}_nC_1 - (0.65)^2(0.35)^{10}_nC_2 - (0.65)^{12}(0.35)^{10}_nC_2 - (0.65)^{12}$ | A1 | Correct unsimplified answer
$= 1 - 0.0065359 = 0.993$ | A1 [3] | Accept 0.994 from correct working only
(ii) $1 - (0.87)^n > 0.95$ | M1 | Equality or inequality in (0.87 or 0.78 or 0.35), power $n$ or $n - 1$, 0.95 or 0.05
$0.05 > (0.87)^n$ | M1 | Attempt to solve an equation with a power in (can be implied)
$n = 22$ | A1 [3] | Correct answer3 In Restaurant Bijoux 13\% of customers rated the food as 'poor', 22\% of customers rated the food as 'satisfactory' and $65 \%$ rated it as 'good'. A random sample of 12 customers who went for a meal at Restaurant Bijoux was taken.\\
(i) Find the probability that more than 2 and fewer than 12 of them rated the food as 'good'.
On a separate occasion, a random sample of $n$ customers who went for a meal at the restaurant was taken.\\
(ii) Find the smallest value of $n$ for which the probability that at least 1 person will rate the food as 'poor' is greater than 0.95.
\hfill \mbox{\textit{CAIE S1 2012 Q3 [6]}}