CAIE S2 2018 November — Question 1 3 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2018
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeSingle time period probability
DifficultyEasy -1.2 This is a straightforward application of the Poisson distribution requiring only calculator/table lookup of cumulative probabilities and basic probability arithmetic (P(2≤X<5) = P(X≤4) - P(X≤1)). It involves no problem-solving, conceptual depth, or multi-step reasoning—purely routine calculation with a standard distribution.
Spec5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities
1 The random variable \(X\) has the distribution \(\operatorname { Po } ( 2.3 )\). Find \(\mathrm { P } ( 2 \leq X < 5 )\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(e^{-2.3}\left(\frac{2.3^2}{2} + \frac{2.3^3}{3!} + \frac{2.3^4}{4!}\right)\)M2 M1 for one term wrong or one end error or \(1 - P(2, 3, 4)\)
\(= 0.585\) (3 sf)A1
3 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{-2.3}\left(\frac{2.3^2}{2} + \frac{2.3^3}{3!} + \frac{2.3^4}{4!}\right)$ | **M2** | M1 for one term wrong or one end error or $1 - P(2, 3, 4)$ |
| $= 0.585$ (3 sf) | **A1** | |
| | **3** | |

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1 The random variable $X$ has the distribution $\operatorname { Po } ( 2.3 )$. Find $\mathrm { P } ( 2 \leq X < 5 )$.\\

\hfill \mbox{\textit{CAIE S2 2018 Q1 [3]}}
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