| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Moderate -0.8 This is a straightforward Poisson distribution question requiring only standard recall and routine calculations. Part (i) asks for a simple condition (independence/randomness), while parts (ii)-(iv) involve direct application of the Poisson formula with appropriate parameter adjustments and normal approximation—all standard S2 techniques with no problem-solving 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Customers arrive independently or randomly | B1 [1] | In context. Allow "singly" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{-6} \times \dfrac{6^5}{5!}\) | M1 | Poisson \(P(5)\), allow any mean |
| \(= 0.161\) (3 sf) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda = 2.4\) | B1 | |
| \(e^{-2}\left(1 + 2.4 + \dfrac{2.4^2}{2!}\right)\) | M1 | Poisson \(P(0,1,2)\), allow their mean; allow one end error |
| \(= 0.570\) (3 sf) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N(24, 24)\) | B1 | Stated or implied |
| \(\dfrac{295 - 24}{\sqrt{24}} (= 1.123)\) | M1 | Allow with wrong or no cc and/or no \(\sqrt{\phantom{x}}\); Correct area |
| \(\Phi(1.123)\) | M1 | |
| \(= 0.869\) (3 sf) | A1 [4] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Customers arrive independently or randomly | B1 [1] | In context. Allow "singly" |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-6} \times \dfrac{6^5}{5!}$ | M1 | Poisson $P(5)$, allow any mean |
| $= 0.161$ (3 sf) | A1 [2] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 2.4$ | B1 | |
| $e^{-2}\left(1 + 2.4 + \dfrac{2.4^2}{2!}\right)$ | M1 | Poisson $P(0,1,2)$, allow their mean; allow one end error |
| $= 0.570$ (3 sf) | A1 [3] | |
### Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N(24, 24)$ | B1 | Stated or implied |
| $\dfrac{295 - 24}{\sqrt{24}} (= 1.123)$ | M1 | Allow with wrong or no cc and/or no $\sqrt{\phantom{x}}$; Correct area |
| $\Phi(1.123)$ | M1 | |
| $= 0.869$ (3 sf) | A1 [4] | |
---6 Customers arrive at an enquiry desk at a constant average rate of 1 every 5 minutes.\\
(i) State one condition for the number of customers arriving in a given period to be modelled by a Poisson distribution.
Assume now that a Poisson distribution is a suitable model.\\
(ii) Find the probability that exactly 5 customers will arrive during a randomly chosen 30 -minute period.\\
(iii) Find the probability that fewer than 3 customers will arrive during a randomly chosen 12-minute period.\\
(iv) Find an estimate of the probability that fewer than 30 customers will arrive during a randomly chosen 2-hour period.
\hfill \mbox{\textit{CAIE S2 2011 Q6 [10]}}