| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Combined independent Poisson probabilities |
| Difficulty | Standard +0.8 This question requires understanding when to approximate binomial to Poisson, combining independent Poisson distributions, and solving an inequality involving complementary probability with combined parameters. The multi-step reasoning across two parts, particularly the algebraic manipulation in part (ii) to find n, elevates this above standard S2 questions but remains within typical Further Maths scope. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\lambda_A = np = 0.022 \times 55 = 1.21\), \(\lambda_B = 0.058 \times 55 = 3.19\), total \(\lambda = 4.4\) | M1, A1 [4] | Two different \(np\) (can be implied); Correct total 4.4 (or all method: 6 correct combinations 0.0, 1.0 etc stated and used) |
| \(P(\text{more than 2}) = 1 - P(0, 1, 2) = 1 - e^{-4.4}\left(1 + 4.4 + \frac{4.4^2}{2!}\right) = 1 - 0.185 = 0.815\) | M1, A1 | Finding 1 − P(0, 1, 2), Poisson, any mean, allow one end error. (Or combinations method – use at least 4 and find 1 − P(≤2) ) |
| (ii) \(\lambda = 0.08n\) | B1 | Correct \(\lambda\) |
| \(P(\text{at least 1 stained tablecloth}) = 1 - P(0) = 1 - e^{-0.08n} > 0.99\) | M1 | Equation of correct form relating their \(\lambda\) and 0.99 |
| \(0.01 > e^{-0.08n}\) | M1 | Valid attempt to solve equation of correct form by logs or trial and error |
| \(n > 57.6\), least value of \(n = 58\) | A1 [4] | Correct answer (SR Accept use of Binomial leading to \(n = 57\)) |
**(i)** $\lambda_A = np = 0.022 \times 55 = 1.21$, $\lambda_B = 0.058 \times 55 = 3.19$, total $\lambda = 4.4$ | M1, A1 [4] | Two different $np$ (can be implied); Correct total 4.4 (or all method: 6 correct combinations 0.0, 1.0 etc stated and used)
$P(\text{more than 2}) = 1 - P(0, 1, 2) = 1 - e^{-4.4}\left(1 + 4.4 + \frac{4.4^2}{2!}\right) = 1 - 0.185 = 0.815$ | M1, A1 | Finding 1 − P(0, 1, 2), Poisson, any mean, allow one end error. (Or combinations method – use at least 4 and find 1 − P(≤2) )
**(ii)** $\lambda = 0.08n$ | B1 | Correct $\lambda$
$P(\text{at least 1 stained tablecloth}) = 1 - P(0) = 1 - e^{-0.08n} > 0.99$ | M1 | Equation of correct form relating their $\lambda$ and 0.99
$0.01 > e^{-0.08n}$ | M1 | Valid attempt to solve equation of correct form by logs or trial and error
$n > 57.6$, least value of $n = 58$ | A1 [4] | Correct answer (SR Accept use of Binomial leading to $n = 57$)6 In restaurant $A$ an average of 2.2\% of tablecloths are stained and, independently, in restaurant $B$ an average of 5.8\% of tablecloths are stained.\\
(i) Random samples of 55 tablecloths are taken from each restaurant. Use a suitable Poisson approximation to find the probability that a total of more than 2 tablecloths are stained.\\
(ii) Random samples of $n$ tablecloths are taken from each restaurant. The probability that at least one tablecloth is stained is greater than 0.99 . Find the least possible value of $n$.
\hfill \mbox{\textit{CAIE S2 2010 Q6 [8]}}