| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question requiring only scaling the rate parameter (λ = 1.4 × 2.5 = 3.5 for part i, λ = 1.4 × 672 = 940.8 for part ii) and calculating probabilities using standard formulas or tables. Part (ii) may require normal approximation, but this is a standard technique at this level. No conceptual difficulty or novel insight required. |
| 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) | \(\lambda\ (= 1.4 \times 2.5) = 3.5\); \(1 - e^{-3.5}\!\left(1 + 3.5 + \frac{3.5^2}{2} + \frac{3.5^3}{3!}\right)\); \(= 0.463\) (3 sf) | B1, M1, A1 [3] |
| (ii) | \((\lambda = 672 \times 1.4 = 940.8)\); \(N(940.8,\ 940.8)\); \(\frac{999.5 - 940.8}{\sqrt{940.8}}\ (= 1.914)\); \(\Phi(1.914)\); \(= 0.972\) (3 sf) | B1, M1, M1, A1 [4] |
## Question 6:
**(i)** | $\lambda\ (= 1.4 \times 2.5) = 3.5$; $1 - e^{-3.5}\!\left(1 + 3.5 + \frac{3.5^2}{2} + \frac{3.5^3}{3!}\right)$; $= 0.463$ (3 sf) | B1, M1, A1 [3] | Any $\lambda$, allow one end error |
**(ii)** | $(\lambda = 672 \times 1.4 = 940.8)$; $N(940.8,\ 940.8)$; $\frac{999.5 - 940.8}{\sqrt{940.8}}\ (= 1.914)$; $\Phi(1.914)$; $= 0.972$ (3 sf) | B1, M1, M1, A1 [4] | Seen or implied; allow with wrong or no cc, no sd/var mixes |
---6 Calls arrive at a helpdesk randomly and at a constant average rate of 1.4 calls per hour. Calculate the probability that there will be\\
(i) more than 3 calls in $2 \frac { 1 } { 2 }$ hours,\\
(ii) fewer than 1000 calls in four weeks ( 672 hours).
\hfill \mbox{\textit{CAIE S2 2013 Q6 [7]}}