OCR S2 2013 June — Question 2 4 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2013
SessionJune
Marks4
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TopicApproximating the Poisson to the Normal distribution
TypeScaled Poisson over time period
DifficultyStandard +0.3 This is a straightforward application of the normal approximation to the Poisson distribution with a large parameter. Students must scale the Poisson parameter for 40 seconds (giving λ = 2×10⁶), recognize that the large λ justifies normal approximation, apply continuity correction, and perform a standard z-score calculation. While the scientific notation and large numbers may appear intimidating, the question requires only routine application of a standard technique with no conceptual challenges or novel problem-solving.
Spec5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY)
2 The number of neutrinos that pass through a certain region in one second is a random variable with the distribution \(\operatorname { Po } \left( 5 \times 10 ^ { 4 } \right)\). Use a suitable approximation to calculate the probability that the number of neutrinos passing through the region in 40 seconds is less than \(1.999 \times 10 ^ { 6 }\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\text{Po}(2 \times 10^6)\)M1 \(N(\text{their } 40\lambda)\)
\(\approx N(2 \times 10^6,\ 2 \times 10^6)\)A1 Both parameters correct, allow \(\sqrt{\phantom{x}}\) here
\(\Phi\!\left(\dfrac{1998999.5 - 2\times10^6}{\sqrt{2\times10^6}}\right) = \Phi(-0.70746)\)A1 Standardise, mean \(40\lambda\), sd \(\sqrt{40\lambda}\) (*not* \(40\lambda\)). Correct cc must be seen for this A1
\(= 0.2396\)A1 Answer, a.r.t. 0.240 (no cc: M1A1A0A1). NB: no cc gives \(\Phi(-0.7071)\), 0.23975; wrong cc gives \(\Phi(-0.70675)\), 0.23986
Total: 4 
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Po}(2 \times 10^6)$ | M1 | $N(\text{their } 40\lambda)$ |
| $\approx N(2 \times 10^6,\ 2 \times 10^6)$ | A1 | Both parameters correct, allow $\sqrt{\phantom{x}}$ here |
| $\Phi\!\left(\dfrac{1998999.5 - 2\times10^6}{\sqrt{2\times10^6}}\right) = \Phi(-0.70746)$ | A1 | Standardise, mean $40\lambda$, sd $\sqrt{40\lambda}$ (*not* $40\lambda$). Correct cc must be seen for this A1 |
| $= 0.2396$ | A1 | Answer, a.r.t. 0.240 (no cc: M1A1A0A1). NB: no cc gives $\Phi(-0.7071)$, 0.23975; wrong cc gives $\Phi(-0.70675)$, 0.23986 |
| | **Total: 4** | |

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2 The number of neutrinos that pass through a certain region in one second is a random variable with the distribution $\operatorname { Po } \left( 5 \times 10 ^ { 4 } \right)$. Use a suitable approximation to calculate the probability that the number of neutrinos passing through the region in 40 seconds is less than $1.999 \times 10 ^ { 6 }$.

\hfill \mbox{\textit{OCR S2 2013 Q2 [4]}}
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