Edexcel S2 — Question 2 9 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeSingle time period probability
DifficultyModerate -0.8 This is a straightforward Poisson distribution question requiring only recognition of the appropriate model (part a), basic conceptual understanding of the parameter (part b), and direct application of standard Poisson probability formulas (parts c-d). All calculations are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure and cumulative probability calculation.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda

An ornithologist believes that on average 4.2 different species of bird will visit a bird table in a rural garden when 50 g of breadcrumbs are spread on it.
  1. Suggest a suitable distribution for modelling the number of species that visit a bird table meeting these criteria. [1 mark]
  2. Explain why the parameter used with this model may need to be changed if
    1. 50 g of nuts are used instead of breadcrumbs,
    2. 100g of breadcrumbs are used.
    [2 marks]
A bird table in a rural garden has 50 g of breadcrumbs spread on it. Find the probability that
  1. exactly 6 different species visit the table, [2 marks]
  2. more than 2 different species visit the table. [4 marks]

AnswerMarks
(a) Poisson (with \(\lambda = 4.2\))B1
(b)(i) e.g. may be more or less species that like nutsB1
(b)(ii) e.g. will last longer so may get more species visitingB1
(c) Let \(X =\) no. of species that visit \(\therefore X \sim \text{Po}(4.2)\); \(P(X = 6) = \frac{e^{-4.2} \times 4.2^6}{6!} = 0.1143\) (4sf)M1 A1
(d) \(P(X > 2) = 1 - P(X \leq 2) = 1 - e^{-4.2}(1 + 4.2 + \frac{4.2^2}{2}) = 1 - 0.2102 = 0.7898\) (4sf)M1 M1 A1
(a) Poisson (with $\lambda = 4.2$) | B1 |

(b)(i) e.g. may be more or less species that like nuts | B1 |

(b)(ii) e.g. will last longer so may get more species visiting | B1 |

(c) Let $X =$ no. of species that visit $\therefore X \sim \text{Po}(4.2)$; $P(X = 6) = \frac{e^{-4.2} \times 4.2^6}{6!} = 0.1143$ (4sf) | M1 A1 |

(d) $P(X > 2) = 1 - P(X \leq 2) = 1 - e^{-4.2}(1 + 4.2 + \frac{4.2^2}{2}) = 1 - 0.2102 = 0.7898$ (4sf) | M1 M1 A1 |

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An ornithologist believes that on average 4.2 different species of bird will visit a bird table in a rural garden when 50 g of breadcrumbs are spread on it.

\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable distribution for modelling the number of species that visit a bird table meeting these criteria. [1 mark]

\item Explain why the parameter used with this model may need to be changed if
\begin{enumerate}[label=(\roman*)]
\item 50 g of nuts are used instead of breadcrumbs,
\item 100g of breadcrumbs are used.
\end{enumerate}
[2 marks]
\end{enumerate}

A bird table in a rural garden has 50 g of breadcrumbs spread on it.

Find the probability that

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item exactly 6 different species visit the table, [2 marks]
\item more than 2 different species visit the table. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q2 [9]}}