| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Scaled time period sums |
| Difficulty | Moderate -0.8 This is a straightforward application of the Poisson distribution with clearly stated conditions. Parts (i)-(iv) are routine calculations requiring only direct use of the Poisson formula or tables. Part (v) adds a minor complication of combining two Poisson distributions, but this is a standard technique. No novel insight or complex problem-solving is required—just methodical application of learned procedures. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (i) | Poisson(6.1) |
| [1] | 3.3 | Do not insist on the parameter shown |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | 6.1 | B1 |
| [1] | 1.1 | |
| (iii) | P(≥ 6) = 0.570 (0.57024...) | B1 |
| [1] | 1.1 | Accept 0.57 |
| Answer | Marks |
|---|---|
| (iv) | New mean = 36.6 |
| P(≥ 36) = 0.562 (0.56157...) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1a | |
| 1.1 | For mean | 36.6 or 6.1×6 seen |
| Answer | Marks |
|---|---|
| (v) | New mean = 9.5 |
| Answer | Marks |
|---|---|
| P(10 ≤ X ≤ 15) = 0.445 (0.44470...) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1a | |
| 1.1 | For mean and one correct probability | Accept 0.44 www |
Question 1:
1 | (i) | Poisson(6.1) | B1
[1] | 3.3 | Do not insist on the parameter shown | Accept Po(6.1)
Do not accept Po
(ii) | 6.1 | B1
[1] | 1.1
(iii) | P(≥ 6) = 0.570 (0.57024...) | B1
[1] | 1.1 | Accept 0.57
Accept 0.5702
(iv) | New mean = 36.6
P(≥ 36) = 0.562 (0.56157...) | M1
A1
[2] | 1.1a
1.1 | For mean | 36.6 or 6.1×6 seen
Accept 0.56, 0.5616
(v) | New mean = 9.5
P(≤ 9) = 0.5218... P(≤ 15) = 0.9665...
P(10 ≤ X ≤ 15) = 0.445 (0.44470...) | M1
A1
[2] | 1.1a
1.1 | For mean and one correct probability | Accept 0.44 www
Accept 0.44471 Over a period of time, radioactive substances decay into other substances. During this decay a Geiger counter can be used to detect the number of radioactive particles that the substance emits.
A certain radioactive source is decaying at a constant average rate of 6.1 particles per 10 seconds. The particles are emitted randomly and independently of each other.\\
(i) State a distribution which can be used to model the number of particles emitted by the source in a 10-second period.\\
(ii) State the variance of this distribution.\\
(iii) Find the probability that at least 6 particles are detected in a period of 10 seconds.\\
(iv) Find the probability that at least 36 particles are detected in a period of 60 seconds.\\
(v) Another radioactive source emits particles randomly and independently at a constant average rate of 1.7 particles per 5 seconds. Find the probability that at least 10 but no more than 15 particles are detected altogether from the two sources in a period of 10 seconds.
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2018 Q1 [7]}}