| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Rescale rate then sum Poissons |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard calculations: (i) requires finding P(X<4) using tables/calculator, (ii) uses normal approximation to Poisson (a standard S2 technique), and (iii) applies the sum of independent Poisson variables property. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| 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 probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\lambda = 1.25\) | \(M1\) | For attempting to find new \(\lambda\) and using it |
| \(P(X < 4) = e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2} + \frac{1.25^3}{6}\right)\) | \(M1\) | For summing \(P(0), 1, 2, 3)\) or \(P(0, 1, 2, 3, 4)\) using a Poisson expression |
| \(= 0.962\) | \(A1\) | For correct answer |
| (ii) \(X \sim N(182.5, 182.5)\) | \(B1, M1\) | For correct mean and variance |
| \(P(> 200 \text{ breakdowns}) = 1 - \Phi\left(\frac{200.5 - 182.5}{\sqrt{182.5}}\right)\) | For standardising process with or without continuity correction | |
| \(= 1 - \Phi(1.332)\) | \(A1ft\) | For correct standardising and correct tail |
| \(= 0.0915\) \((0.0914)\) | \(A1\) | For correct answer |
| (iii) \(\lambda = 5\) for phone calls | \(B1\) | |
| \(\lambda = 6.25\) for total | ||
| \(P(X = 4) = e^{-6.25}\left(\frac{6.25^4}{4!}\right)\) | \(M1\) | For summing their two \(\lambda\)s and using a Poisson expression OR alt. method using sep. distributions 5 terms req. |
| \(= 0.123\) | \(A1\) | For correct answer |
**(i)** $\lambda = 1.25$ | $M1$ | For attempting to find new $\lambda$ and using it
$P(X < 4) = e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2} + \frac{1.25^3}{6}\right)$ | $M1$ | For summing $P(0), 1, 2, 3)$ or $P(0, 1, 2, 3, 4)$ using a Poisson expression
$= 0.962$ | $A1$ | For correct answer
**(ii)** $X \sim N(182.5, 182.5)$ | $B1, M1$ | For correct mean and variance
$P(> 200 \text{ breakdowns}) = 1 - \Phi\left(\frac{200.5 - 182.5}{\sqrt{182.5}}\right)$ | | For standardising process with or without continuity correction
$= 1 - \Phi(1.332)$ | $A1ft$ | For correct standardising and correct tail
$= 0.0915$ $(0.0914)$ | $A1$ | For correct answer
**(iii)** $\lambda = 5$ for phone calls | $B1$ |
$\lambda = 6.25$ for total | |
$P(X = 4) = e^{-6.25}\left(\frac{6.25^4}{4!}\right)$ | $M1$ | For summing their two $\lambda$s and using a Poisson expression OR alt. method using sep. distributions 5 terms req.
$= 0.123$ | $A1$ | For correct answer6 Computer breakdowns occur randomly on average once every 48 hours of use.\\
(i) Calculate the probability that there will be fewer than 4 breakdowns in 60 hours of use.\\
(ii) Find the probability that the number of breakdowns in one year (8760 hours) of use is more than 200.\\
(iii) Independently of the computer breaking down, the computer operator receives phone calls randomly on average twice in every 24 -hour period. Find the probability that the total number of phone calls and computer breakdowns in a 60-hour period is exactly 4 .
\hfill \mbox{\textit{CAIE S2 2003 Q6 [10]}}