| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with parameter scaling (4.8 faults per 20m² → 0.6 faults per 2.5m²), combined with basic probability multiplication and a normal approximation to binomial. All parts follow standard procedures with no novel insight required, making it slightly easier than average for S2 level. |
| 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 |
|---|---|---|
| \(1 - e^{-0.6(1+0.6)} = 1 - 0.878 = 0.122\) | M1, B1, A1 (3 marks) | Poisson calc \(1 - P(0,1)\). Correct numerical expression for \(P(0,1)\). Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \((e^{-0.6})(0.97) = 0.532\) | M1, B1 (2 marks) | Multiplying \(P(0)\) for skirt by \(0.97\). Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(F,F) = (1-e^{-0.6})(0.03) = 0.0135\) \((0.01354)\) | M1, A1 (2 marks) | Finding \(P(F,F)\). Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X<3) = e^{-4.062}\left(1+4.062+\frac{4.062^2}{2}\right) = 0.229\) | M1, A1ft, A1 (3 marks) | Using poisson approx, \(\lambda = 300 \times\) their \(P(F,F)\). Correct numerical expression for \(P(X<3)\) with their \(\lambda\). Correct answer |
**Part (i):**
$1 - e^{-0.6(1+0.6)} = 1 - 0.878 = 0.122$ | M1, B1, A1 (3 marks) | Poisson calc $1 - P(0,1)$. Correct numerical expression for $P(0,1)$. Correct answer
**Part (ii):**
$(e^{-0.6})(0.97) = 0.532$ | M1, B1 (2 marks) | Multiplying $P(0)$ for skirt by $0.97$. Correct answer
**Part (iii):**
$P(F,F) = (1-e^{-0.6})(0.03) = 0.0135$ $(0.01354)$ | M1, A1 (2 marks) | Finding $P(F,F)$. Correct answer
**Part (iv):**
$X$ approx $P(300 \times 0.01354) = P(4.062)$
$P(X<3) = e^{-4.062}\left(1+4.062+\frac{4.062^2}{2}\right) = 0.229$ | M1, A1ft, A1 (3 marks) | Using poisson approx, $\lambda = 300 \times$ their $P(F,F)$. Correct numerical expression for $P(X<3)$ with their $\lambda$. Correct answer6 A dressmaker makes dresses for Easifit Fashions. Each dress requires $2.5 \mathrm {~m} ^ { 2 }$ of material. Faults occur randomly in the material at an average rate of 4.8 per $20 \mathrm {~m} ^ { 2 }$.\\
(i) Find the probability that a randomly chosen dress contains at least 2 faults.
Each dress has a belt attached to it to make an outfit. Independently of faults in the material, the probability that a belt is faulty is 0.03 . Find the probability that, in an outfit,\\
(ii) neither the dress nor its belt is faulty,\\
(iii) the dress has at least one fault and its belt is faulty.
The dressmaker attaches 300 randomly chosen belts to 300 randomly chosen dresses. An outfit in which the dress has at least one fault and its belt is faulty is rejected.\\
(iv) Use a suitable approximation to find the probability that fewer than 3 outfits are rejected.
\hfill \mbox{\textit{CAIE S2 2006 Q6 [10]}}