Edexcel S2 2004 June — Question 6 12 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2004
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeSingle period normal approximation - large lambda direct
DifficultyStandard +0.3 This is a straightforward Poisson distribution question requiring standard applications: identifying the model, calculating probabilities using tables/calculator, and applying normal approximation. Part (d) requires recognizing when to use continuity correction, which is slightly above routine but still a standard S2 technique. The question involves no novel problem-solving or conceptual challenges beyond textbook exercises.
Spec5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY)

Minor defects occur in a particular make of carpet at a mean rate of 0.05 per m\(^2\).
  1. Suggest a suitable model for the distribution of the number of defects in this make of carpet. Give a reason for your answer.
A carpet fitter has a contract to fit this carpet in a small hotel. The hotel foyer requires 30 m\(^2\) of this carpet. Find the probability that the foyer carpet contains
  1. exactly 2 defects, [3]
  2. more than 5 defects. [3]
The carpet fitter orders a total of 355 m\(^2\) of the carpet for the whole hotel.
  1. Using a suitable approximation, find the probability that this total area of carpet contains 22 or more defects. [6]

Part (a)
AnswerMarks Guidance
No of defects in carpet area \(a\) sq m is distributed \(Po(0.05a)\). Defects occur at a constant rate, independent, singly, randomlyB1B1, Any 1 B1 (3)
Part (b)
\(X \sim P(30 \times 0.05) = P(1.5)\)
AnswerMarks Guidance
\(P(X = 2) = \frac{e^{-1.5} \times 1.5^2}{2} = 0.2510\)B1, M1A1 (3)
Tables or calc 0.251(0)
Part (c)
AnswerMarks Guidance
\(P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9955 = 0.0045\)M1M1A1 (3)
Strict inequality, 1-0.9955, 0.0045
Part (d)
\(X \sim P(17.75)\)
\(X \sim N(17.75, 17.75)\)
\(P(X \geq 22) = P\left(Z > \frac{21.5 - 17.75}{\sqrt{17.75}}\right)\)
\(= P(Z > 0.89)\)
AnswerMarks Guidance
\(= 0.1867\)B1, B1, M1M1, awrt 0.89, A1, A1 (6)
Standardise, accept 22 or ±0.5
## Part (a)
No of defects in carpet area $a$ sq m is distributed $Po(0.05a)$. Defects occur at a constant rate, independent, singly, randomly | B1B1, Any 1 B1 | (3)

## Part (b)
$X \sim P(30 \times 0.05) = P(1.5)$

$P(X = 2) = \frac{e^{-1.5} \times 1.5^2}{2} = 0.2510$ | B1, M1A1 | (3)

Tables or calc 0.251(0)

## Part (c)
$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9955 = 0.0045$ | M1M1A1 | (3)

Strict inequality, 1-0.9955, 0.0045

## Part (d)
$X \sim P(17.75)$

$X \sim N(17.75, 17.75)$

$P(X \geq 22) = P\left(Z > \frac{21.5 - 17.75}{\sqrt{17.75}}\right)$

$= P(Z > 0.89)$

$= 0.1867$ | B1, B1, M1M1, awrt 0.89, A1, A1 | (6)

Standardise, accept 22 or ±0.5
Minor defects occur in a particular make of carpet at a mean rate of 0.05 per m$^2$.

\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable model for the distribution of the number of defects in this make of carpet. Give a reason for your answer.
\end{enumerate}

A carpet fitter has a contract to fit this carpet in a small hotel. The hotel foyer requires 30 m$^2$ of this carpet. Find the probability that the foyer carpet contains

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item exactly 2 defects, [3]
\item more than 5 defects. [3]
\end{enumerate}

The carpet fitter orders a total of 355 m$^2$ of the carpet for the whole hotel.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Using a suitable approximation, find the probability that this total area of carpet contains 22 or more defects. [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2004 Q6 [12]}}