Edexcel S2 — Question 5 13 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeSingle period normal approximation - scaled period (exact Poisson part)
DifficultyStandard +0.3 This is a standard S2 Poisson distribution question with routine applications: identifying the model, calculating P(X>2), finding probability for multiple pages, and applying normal approximation with continuity correction. All techniques are textbook exercises 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

A textbook contains, on average, 1.2 misprints per page. Assuming that the misprints are randomly distributed throughout the book,
  1. specify a suitable model for \(X\), the random variable representing the number of misprints on a given page. [1 mark]
  2. Find the probability that a particular page has more than 2 misprints. [3 marks]
  3. Find the probability that Chapter 1, with 8 pages, has no misprints at all. [2 marks]
Chapter 2 is longer, at 20 pages.
  1. Use a suitable approximation to find the probability that Chapter 2 has less than ten misprints altogether. Explain what adjustment is necessary when making this approximation. [7 marks]

AnswerMarks Guidance
(a) Poisson: \(X \sim Po(1.2)\)B1
(b) \(P(X > 2) = 1 - e^{-1.2} - 1.2e^{-1.2} - 1.2^2 e^{-1.2}/2! = 0.121\)M1 A1 A1
(c) \(P(X = 0) = e^{-1.2} = 0.301\) \(P(0 \text{ in Ch. 1}) = 0.301^8 = 0.0000677\)M1 A1
(d) Total for Ch. 2 \(\sim Po(24)\) \(N(24, 24)\) Then \(P(X < 10)\)M1 A1
\(= P(X < 9.5) = P\left(Z < \frac{-14.5}{4.90}\right) = P(Z < -2.96) = 0.0015\)M1 A1 M1 A1
Continuity correction needed, from discrete to continuousB1 13 marks total
(a) Poisson: $X \sim Po(1.2)$ | B1 |

(b) $P(X > 2) = 1 - e^{-1.2} - 1.2e^{-1.2} - 1.2^2 e^{-1.2}/2! = 0.121$ | M1 A1 A1 |

(c) $P(X = 0) = e^{-1.2} = 0.301$ $P(0 \text{ in Ch. 1}) = 0.301^8 = 0.0000677$ | M1 A1 |

(d) Total for Ch. 2 $\sim Po(24)$ $N(24, 24)$ Then $P(X < 10)$ | M1 A1 |

$= P(X < 9.5) = P\left(Z < \frac{-14.5}{4.90}\right) = P(Z < -2.96) = 0.0015$ | M1 A1 M1 A1 |

Continuity correction needed, from discrete to continuous | B1 | 13 marks total
A textbook contains, on average, 1.2 misprints per page. Assuming that the misprints are randomly distributed throughout the book,
\begin{enumerate}[label=(\alph*)]
\item specify a suitable model for $X$, the random variable representing the number of misprints on a given page. [1 mark]
\item Find the probability that a particular page has more than 2 misprints. [3 marks]
\item Find the probability that Chapter 1, with 8 pages, has no misprints at all. [2 marks]
\end{enumerate}

Chapter 2 is longer, at 20 pages.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Use a suitable approximation to find the probability that Chapter 2 has less than ten misprints altogether. Explain what adjustment is necessary when making this approximation. [7 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q5 [13]}}