CAIE S2 2020 June — Question 5 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2020
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSum of Poisson processes
TypeProbability of ordered outcome between components
DifficultyStandard +0.3 This is a straightforward application of Poisson distribution properties with standard techniques: part (a) requires summing independent Poissons and enumerating cases, part (b) uses normal approximation to Poisson (a routine S2 technique), and part (c) asks for standard justification (np > 5). The calculations are methodical rather than insightful, making it slightly easier than average for an S2 question.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY)

5 Each week a sports team plays one home match and one away match. In their home matches they score goals at a constant average rate of 2.1 goals per match. In their away matches they score goals at a constant average rate of 0.8 goals per match. You may assume that goals are scored at random times and independently of one another.
  1. A week is chosen at random.
    1. Find the probability that the team scores a total of 4 goals in their two matches.
    2. Find the probability that the team scores a total of 4 goals, with more goals scored in the home match than in the away match.
  2. Use a suitable approximating distribution to find the probability that the team scores fewer than 25 goals in 10 randomly chosen weeks.
  3. Justify the use of the approximating distribution used in part (b).

Question 5:
Part 5(a)(i):
AnswerMarks Guidance
AnswerMark Guidance
\(e^{-2.9} \times \dfrac{2.9^4}{4!}\)M1
\(0.162\) (3 sf)A1
Part 5(a)(ii):
AnswerMarks Guidance
AnswerMark Guidance
\(e^{-2.1} \times \dfrac{2.1^4}{4!} \times e^{-0.8} + e^{-2.1} \times \dfrac{2.1^3}{3!} \times e^{-0.8} \times 0.8\)B1 B1 for either expression correct; M1 for \(P(4,0) + P(3,1)\)
M1
\(0.113\) (3 sf)A1
Part 5(b):
AnswerMarks Guidance
AnswerMark Guidance
\(N(29,\ 29)\)M1
\(\dfrac{24.5 - 29}{\sqrt{29}}\) \((= -0.83563)\)M1
\(1 - \Phi(\text{"0.836"})\)M1
\(0.202\) (3 sf)A1
Part 5(c):
AnswerMarks Guidance
AnswerMark Guidance
\(29\) is large or \(29 > 15\)B1
## Question 5:

**Part 5(a)(i):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-2.9} \times \dfrac{2.9^4}{4!}$ | M1 | |
| $0.162$ (3 sf) | A1 | |

**Part 5(a)(ii):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-2.1} \times \dfrac{2.1^4}{4!} \times e^{-0.8} + e^{-2.1} \times \dfrac{2.1^3}{3!} \times e^{-0.8} \times 0.8$ | B1 | B1 for either expression correct; M1 for $P(4,0) + P(3,1)$ |
| | M1 | |
| $0.113$ (3 sf) | A1 | |

**Part 5(b):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $N(29,\ 29)$ | M1 | |
| $\dfrac{24.5 - 29}{\sqrt{29}}$ $(= -0.83563)$ | M1 | |
| $1 - \Phi(\text{"0.836"})$ | M1 | |
| $0.202$ (3 sf) | A1 | |

**Part 5(c):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $29$ is large or $29 > 15$ | B1 | |

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5 Each week a sports team plays one home match and one away match. In their home matches they score goals at a constant average rate of 2.1 goals per match. In their away matches they score goals at a constant average rate of 0.8 goals per match. You may assume that goals are scored at random times and independently of one another.
\begin{enumerate}[label=(\alph*)]
\item A week is chosen at random.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the team scores a total of 4 goals in their two matches.
\item Find the probability that the team scores a total of 4 goals, with more goals scored in the home match than in the away match.
\end{enumerate}\item Use a suitable approximating distribution to find the probability that the team scores fewer than 25 goals in 10 randomly chosen weeks.
\item Justify the use of the approximating distribution used in part (b).
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2020 Q5 [10]}}