CAIE S2 2015 November — Question 1 4 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2015
SessionNovember
Marks4
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Mark schemeDownload PDF ↗
TopicSum of Poisson processes
TypeBasic sum of two Poissons
DifficultyStandard +0.3 This requires knowing that independent Poisson distributions sum to another Poisson distribution, adjusting rates for the 6-month period (halving the annual rates), then calculating P(2 ≤ X ≤ 3) using standard Poisson probability formulas. It's a straightforward application of Poisson properties with routine calculations, slightly above average due to the multi-step setup but well within standard S2 material.
Spec5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02n Sum of Poisson variables: is Poisson
1 Failures of two computers occur at random and independently. On average the first computer fails 1.2 times per year and the second computer fails 2.3 times per year. Find the probability that the total number of failures by the two computers in a 6-month period is more than 1 and less than 4 .
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\lambda = (1.2 + 2.3) \div 2 = 1.75\)M1 Attempt combined mean, allow \(1.2 + 2.3\)
Correct meanA1 Correct mean
\(e^{-1.75}\left(\frac{1.75^2}{2} + \frac{1.75^3}{3!}\right)\)M1 Allow incorrect mean. Allow end errors (1 and/or 4)
\(= 0.421\) (3 sf)A1 [4] 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = (1.2 + 2.3) \div 2 = 1.75$ | M1 | Attempt combined mean, allow $1.2 + 2.3$ |
| Correct mean | A1 | Correct mean |
| $e^{-1.75}\left(\frac{1.75^2}{2} + \frac{1.75^3}{3!}\right)$ | M1 | Allow incorrect mean. Allow end errors (1 and/or 4) |
| $= 0.421$ (3 sf) | A1 [4] | |

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1 Failures of two computers occur at random and independently. On average the first computer fails 1.2 times per year and the second computer fails 2.3 times per year. Find the probability that the total number of failures by the two computers in a 6-month period is more than 1 and less than 4 .

\hfill \mbox{\textit{CAIE S2 2015 Q1 [4]}}
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