Edexcel S2 — Question 5 14 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeSingle period normal approximation - large lambda direct
DifficultyStandard +0.3 This is a straightforward application of the Poisson distribution with standard calculations (P(X=0) and P(X>7)) plus a normal approximation for large λ. Part (c) requires basic interpretation, and part (d) is a routine normal approximation with continuity correction. All techniques are standard S2 material with no novel problem-solving required, making it slightly easier than average.
Spec2.04d 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 probabilities

  1. In World War II, the number of V2 missiles that landed on each square mile of London was, on average, \(3 \cdot 5\). Assuming that the hits were randomly distributed throughout London,
    1. suggest a suitable model for the number of hits on each square mile, giving a suitable value for any parameters.
    2. calculate the probability that a particular square mile received
      1. no hits,
      2. more than 7 hits.
    3. State, with a reason, whether the model is likely to be accurate.
    In contrast, the number of bombs weighing more than 1 ton landing on each square mile was 45 .
  2. Use a suitable approximation to find the probability that a randomly selected square mile received more than 60 such bombs. Explain what adjustment must be made when using this approximation.

AnswerMarks
(a) Poisson, \(\text{Po}(3.5)\)B1
(b) (i) \(P(X = 0) = 0.0302\) (from tables)B1
(ii) \(P(X > 7) = 1 - P(X \leq 7) = 1 - 0.9733 = 0.0267\)M1 A1
(c) Might not be random – possibly aimed at specific targetsB1
(d) \(\text{Po}(45) \approx N(45, 45)\)M1 A1
\(P(X > 60) = P(X > 60.5) = P(Z > \frac{15.5/\sqrt{71}}{}) = P(Z > 2.31)\)M1 A1 A1
\(= 1 - 0.9896 = 0.0104\)M1 A1
A continuity correction must be made to convert from discrete Poisson to continuous Normal distributionB2
Total: 14 marks
(a) Poisson, $\text{Po}(3.5)$ | B1 |

(b) (i) $P(X = 0) = 0.0302$ (from tables) | B1 |

(ii) $P(X > 7) = 1 - P(X \leq 7) = 1 - 0.9733 = 0.0267$ | M1 A1 |

(c) Might not be random – possibly aimed at specific targets | B1 |

(d) $\text{Po}(45) \approx N(45, 45)$ | M1 A1 |

$P(X > 60) = P(X > 60.5) = P(Z > \frac{15.5/\sqrt{71}}{}) = P(Z > 2.31)$ | M1 A1 A1 |

$= 1 - 0.9896 = 0.0104$ | M1 A1 |

A continuity correction must be made to convert from discrete Poisson to continuous Normal distribution | B2 |

**Total: 14 marks**

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\begin{enumerate}
  \item In World War II, the number of V2 missiles that landed on each square mile of London was, on average, $3 \cdot 5$. Assuming that the hits were randomly distributed throughout London,\\
(a) suggest a suitable model for the number of hits on each square mile, giving a suitable value for any parameters.\\
(b) calculate the probability that a particular square mile received\\
(i) no hits,\\
(ii) more than 7 hits.\\
(c) State, with a reason, whether the model is likely to be accurate.
\end{enumerate}

In contrast, the number of bombs weighing more than 1 ton landing on each square mile was 45 .\\
(d) Use a suitable approximation to find the probability that a randomly selected square mile received more than 60 such bombs. Explain what adjustment must be made when using this approximation.\\

\hfill \mbox{\textit{Edexcel S2  Q5 [14]}}