| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum n for P(X≥k) threshold |
| Difficulty | Moderate -0.3 Part (i) is pure recall of Poisson conditions (independence/constant rate). Parts (ii)-(iv) are standard Poisson calculations from S2 syllabus: direct probability, normal approximation, and solving λt equation. Routine application of well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Customers arrive independently | B1 | Valid reason in context, allow "random" |
| (ii) \(1 - 0.9921 = 0.0079\) | M1, A1 | Poisson tables, "1 −", or correct formula ± 1 term. Answer, a.r.t. 0.008 [\(1 - 0.9384 = 0.0606\): M1A0] |
| (iii) \(N(48, 48)\); \(z = \frac{55.5 - 48}{\sqrt{48}} = 1.0825\) | B1, B1, M1 dep | Normal, mean 48. Variance or SD same as mean√. Standardise, wrong or no cc, \(\mu = \lambda\) |
| \(1 - \Phi(1.0825) = 0.1394\) | dep M1, A1 | Use tables, answer < 0.5 |
| (iv) \(e^{-\lambda} < 0.02\); \(\lambda > -\ln 0.02 = 3.9912\) | M1, M1 | Correct formula for \(P(0 \mid \lambda = 4)\) at least. In used. OR \(\lambda = 3.9\) at least by T & I. 3.91(2) seen OR \(\lambda = 3.91\) at least by T & I |
| \(0.4t = 3.912\): \(t = 9.78\) minutes; \(t = 9\) minutes \(47\) seconds | A1, M1, A1 | Divide \(\lambda\) by 0.4 or multiply by 150, any distribution. 587 seconds ± 1 sec [inequalities not needed] |
(i) Customers arrive independently | B1 | Valid reason in context, allow "random"
(ii) $1 - 0.9921 = 0.0079$ | M1, A1 | Poisson tables, "1 −", or correct formula ± 1 term. Answer, a.r.t. 0.008 [$1 - 0.9384 = 0.0606$: M1A0]
(iii) $N(48, 48)$; $z = \frac{55.5 - 48}{\sqrt{48}} = 1.0825$ | B1, B1, M1 dep | Normal, mean 48. Variance or SD same as mean√. Standardise, wrong or no cc, $\mu = \lambda$
$1 - \Phi(1.0825) = 0.1394$ | dep M1, A1 | Use tables, answer < 0.5
(iv) $e^{-\lambda} < 0.02$; $\lambda > -\ln 0.02 = 3.9912$ | M1, M1 | Correct formula for $P(0 \mid \lambda = 4)$ at least. In used. OR $\lambda = 3.9$ at least by T & I. 3.91(2) seen OR $\lambda = 3.91$ at least by T & I
$0.4t = 3.912$: $t = 9.78$ minutes; $t = 9$ minutes $47$ seconds | A1, M1, A1 | Divide $\lambda$ by 0.4 or multiply by 150, any distribution. 587 seconds ± 1 sec [inequalities not needed]
---6 Customers arrive at a post office at a constant average rate of 0.4 per minute.\\
(i) State an assumption needed to model the number of customers arriving in a given time interval by a Poisson distribution.
Assuming that the use of a Poisson distribution is justified,\\
(ii) find the probability that more than 2 customers arrive in a randomly chosen 1 -minute interval,\\
(iii) use a suitable approximation to calculate the probability that more than 55 customers arrive in a given two-hour interval,\\
(iv) calculate the smallest time for which the probability that no customers arrive in that time is less than 0.02 , giving your answer to the nearest second.
\hfill \mbox{\textit{OCR S2 2006 Q6 [14]}}