Edexcel S2 — Question 3 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson with binomial combination
DifficultyModerate -0.3 This is a straightforward application of Poisson distribution with standard bookwork. Part (a) requires stating standard Poisson conditions and calculating λ = 180/40 = 4.5. Parts (b) and (c) involve direct use of Poisson probability formulas and binomial distribution respectively, with no problem-solving insight required. The calculations are routine for S2 level, making this slightly easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02i 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 electrician records the number of repairs of different types of appliances that he makes each day. His records show that over 40 working days he repaired a total of 180 CD players.
  1. Explain why a Poisson distribution may be suitable for modelling the number of CD players he repairs each day and find the parameter for this distribution. [4 marks]
  2. Find the probability that on one particular day he repairs
    1. no CD players,
    2. more than 6 CD players. [3 marks]
  3. Find the probability that over 10 working days he will repair more than 6 CD players on exactly 3 of the days. [3 marks]

Part (a)
AnswerMarks
e.g. requests for repairs likely to occur singly, at random and at a constant rateB3
\(\lambda = \frac{180}{40} = 4.5\)A1
Part (b)
AnswerMarks
let \(X\) = no. of repairs per day \(\therefore X \sim \text{Po}(4.5)\)A1
(i) \(P(X = 0) = 0.0111\)M1 A1
(ii) \(P(X > 6) = 1 - P(X \le 6) = 1 - 0.8311 = 0.1689\)M1 A1
Part (c)
AnswerMarks Guidance
let \(Y\) = no. of days he repairs more than 6 \(\therefore Y \sim B(10, 0.1689)\)M1
\(P(Y = 3) = {}^{10}C_3(0.1689)^3(0.8311)^7 = 0.158\) (3sf)M1 A1 (10)
**Part (a)**
e.g. requests for repairs likely to occur singly, at random and at a constant rate | B3 |

$\lambda = \frac{180}{40} = 4.5$ | A1 |

**Part (b)**
let $X$ = no. of repairs per day $\therefore X \sim \text{Po}(4.5)$ | A1 |

(i) $P(X = 0) = 0.0111$ | M1 A1 |

(ii) $P(X > 6) = 1 - P(X \le 6) = 1 - 0.8311 = 0.1689$ | M1 A1 |

**Part (c)**
let $Y$ = no. of days he repairs more than 6 $\therefore Y \sim B(10, 0.1689)$ | M1 |

$P(Y = 3) = {}^{10}C_3(0.1689)^3(0.8311)^7 = 0.158$ (3sf) | M1 A1 | (10)

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An electrician records the number of repairs of different types of appliances that he makes each day. His records show that over 40 working days he repaired a total of 180 CD players.

\begin{enumerate}[label=(\alph*)]
\item Explain why a Poisson distribution may be suitable for modelling the number of CD players he repairs each day and find the parameter for this distribution. [4 marks]

\item Find the probability that on one particular day he repairs
\begin{enumerate}[label=(\roman*)]
\item no CD players,
\item more than 6 CD players. [3 marks]
\end{enumerate}

\item Find the probability that over 10 working days he will repair more than 6 CD players on exactly 3 of the days. [3 marks]
\end{enumerate}

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