OCR FS1 AS 2018 March — Question 6 7 marks

Exam BoardOCR
ModuleFS1 AS (Further Statistics 1 AS)
Year2018
SessionMarch
Marks7
TopicPoisson distribution
TypePoisson parameter from given probability
DifficultyChallenging +1.2 This question requires understanding of the mode of a Poisson distribution and setting up inequalities P(R=7) ≥ P(R=6) and P(R=7) ≥ P(R=8), then solving algebraically to find 7 ≤ λ ≤ 8. While it involves multiple steps and algebraic manipulation of Poisson probabilities, it's a standard Further Maths Statistics question testing a well-defined property with a clear method, making it moderately above average difficulty.
Spec5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!

6 The discrete random variable \(R\) has the distribution \(\operatorname { Po } ( \lambda )\).
Use an algebraic method to find the range of values of \(\lambda\) for which the single most likely value of \(R\) is 7. [7]

AnswerMarks Guidance
\(P(R = 7) > P(R = 6) \Rightarrow e^{-\lambda}\frac{\lambda^7}{7!} > e^{-\lambda}\frac{\lambda^6}{6!}\)M1\* 3.1a
\(6! \lambda^7 > 7! \lambda^6\)A1 1.1
\(\Rightarrow \lambda > 7\)M1 1.1a
A11.1 One correct limit
\(P(R = 7) > P(R = 8) \Rightarrow e^{-\lambda}\frac{\lambda^7}{7!} > e^{-\lambda}\frac{\lambda^8}{8!}\)depM1 2.1
\(\Rightarrow \lambda < 8\)A1 1.1
\(7 < \lambda < 8\)A1 2.2a
$P(R = 7) > P(R = 6) \Rightarrow e^{-\lambda}\frac{\lambda^7}{7!} > e^{-\lambda}\frac{\lambda^6}{6!}$ | **M1\*** | 3.1a | One correct Poisson prob seen

$6! \lambda^7 > 7! \lambda^6$ | **A1** | 1.1 | Correct inequality

$\Rightarrow \lambda > 7$ | **M1** | 1.1a | Reduce to linear equation in $\lambda$

| **A1** | 1.1 | One correct limit | Allow $=$

$P(R = 7) > P(R = 8) \Rightarrow e^{-\lambda}\frac{\lambda^7}{7!} > e^{-\lambda}\frac{\lambda^8}{8!}$ | **depM1** | 2.1 | Second inequality

$\Rightarrow \lambda < 8$ | **A1** | 1.1 | Second correct limit | Allow $=$ only if $\lambda$ later

$7 < \lambda < 8$ | **A1** | 2.2a | Correct double inequality

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6 The discrete random variable $R$ has the distribution $\operatorname { Po } ( \lambda )$.\\
Use an algebraic method to find the range of values of $\lambda$ for which the single most likely value of $R$ is 7. [7]

\hfill \mbox{\textit{OCR FS1 AS 2018 Q6 [7]}}