| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Basic sum of two Poissons |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution properties: recognizing that independent Poisson variables sum to another Poisson, calculating parameters (5×0.21 + 5×0.24 = 2.25), and finding P(X≥2) = 1-P(X≤1). The assumption about independence is standard bookwork. Slightly above average due to the two-distribution setup, but still routine for S3 level. |
| Spec | 5.02i Poisson distribution: random events model5.02n Sum of Poisson variables: is Poisson |
1 The numbers of minor flaws that occur on reels of copper wire and reels of steel wire have Poisson distributions with means 0.21 per metre and 0.24 per metre respectively. One length of 5 m is cut from each reel.\\
(i) Calculate the probability that the total number of flaws on these two lengths of wire is at least 2 .\\
(ii) State one assumption needed in the calculation.
\hfill \mbox{\textit{OCR S3 2010 Q1 [5]}}