Edexcel S2 — Question 3 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson hypothesis test
DifficultyStandard +0.3 This is a straightforward S2 question testing standard Poisson distribution concepts: calculating mean/variance from a frequency table (routine), recognizing mean≈variance as evidence for Poisson (textbook knowledge), and performing a one-tailed hypothesis test. All steps are procedural with no novel insight required, making it slightly easier than average A-level maths.
Spec5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.05a Sample mean distribution: central limit theorem

A secretarial agency carefully assesses the work of a new recruit, with the following results after 150 pages:
No of errors0123456
No of pages163841291772
  1. Find the mean and variance of the number of errors per page. [4 marks]
  2. Explain how these results support the idea that the number of errors per page follows a Poisson distribution. [1 mark]
  3. After two weeks at the agency, the secretary types a fresh piece of work, six pages long, which is found to contain 15 errors. The director suspects that the secretary was trying especially hard during the early period and that she is now less conscientious. Using a Poisson distribution with the mean found in part (a), test this hypothesis at the 5% significance level. [5 marks]

AnswerMarks
(a) Mean \(= 322/150 = 2.15\), variance \(= 982/150 - 2.147^2 = 1.94\)M1 A1 M1 A1
(b) Mean = Variance, which suggests PoissonB1
(c) Assuming \(\lambda = 2.14666\), no. of errors in 6 pages is \(\text{Po}(12.88)\)B1 B1
Then \(P(X \geq 15) - P(X = 15) = 0.0868 > 5\%\), so do not reject \(H_0\)M1 A1 A1
Total: 10 marks
(a) Mean $= 322/150 = 2.15$, variance $= 982/150 - 2.147^2 = 1.94$ | M1 A1 M1 A1 |

(b) Mean = Variance, which suggests Poisson | B1 |

(c) Assuming $\lambda = 2.14666$, no. of errors in 6 pages is $\text{Po}(12.88)$ | B1 B1 |

Then $P(X \geq 15) - P(X = 15) = 0.0868 > 5\%$, so do not reject $H_0$ | M1 A1 A1 |

**Total: 10 marks**

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A secretarial agency carefully assesses the work of a new recruit, with the following results after 150 pages:

\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
No of errors & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
No of pages & 16 & 38 & 41 & 29 & 17 & 7 & 2 \\
\hline
\end{tabular}

\begin{enumerate}[label=(\alph*)]
\item Find the mean and variance of the number of errors per page. [4 marks]
\item Explain how these results support the idea that the number of errors per page follows a Poisson distribution. [1 mark]
\item After two weeks at the agency, the secretary types a fresh piece of work, six pages long, which is found to contain 15 errors.

The director suspects that the secretary was trying especially hard during the early period and that she is now less conscientious. Using a Poisson distribution with the mean found in part (a), test this hypothesis at the 5% significance level. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q3 [10]}}