Edexcel S2 — Question 4 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 Poisson distribution question requiring (a) a standard probability calculation P(X≥2) = 1 - P(X≤1) and (b) a hypothesis test with clearly defined steps. Both parts follow textbook procedures with no novel insight required, though the hypothesis testing component adds slight complexity beyond pure recall.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02i 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

A rugby player scores an average of 0.4 tries per match in which he plays.
  1. Find the probability that he scores 2 or more tries in a match. [5 marks]
The team's coach moves the player to a different position in the team believing he will then score more frequently. In the next five matches he scores 6 tries.
  1. Stating your hypotheses clearly, test at the 5% level of significance whether or not there is evidence of an increase in the mean number of tries the player scores per match as a result of playing in a different position. [5 marks]

AnswerMarks Guidance
(a)let \(X =\) no. of tries per match \(\therefore X \sim \text{Po}(0.4)\) M1
\(P(X \geq 2) = 1 - P(X \leq 1)\)M1
\(= 1 - e^{-0.4}(1 + 0.4)\)M1 A1
\(= 1 - 0.9384 = 0.0616 \text{ (3sf)}\)A1
(b)let \(Y =\) no. of tries per 5 matches \(\therefore Y \sim \text{Po}(2)\) M1
\(H_0: \lambda = 2\) \(H_1: \lambda > 2\)B1
\(P(Y \geq 6) = 1 - P(Y \leq 5) = 1 - 0.9834 = 0.0166\)M1 A1
less than 5% \(\therefore\) significant, evidence of increaseA1 (10)
(a) | let $X =$ no. of tries per match $\therefore X \sim \text{Po}(0.4)$ | M1 |
| $P(X \geq 2) = 1 - P(X \leq 1)$ | M1 |
| $= 1 - e^{-0.4}(1 + 0.4)$ | M1 A1 |
| $= 1 - 0.9384 = 0.0616 \text{ (3sf)}$ | A1 |
(b) | let $Y =$ no. of tries per 5 matches $\therefore Y \sim \text{Po}(2)$ | M1 |
| $H_0: \lambda = 2$ $H_1: \lambda > 2$ | B1 |
| $P(Y \geq 6) = 1 - P(Y \leq 5) = 1 - 0.9834 = 0.0166$ | M1 A1 |
| less than 5% $\therefore$ significant, evidence of increase | A1 | (10)
A rugby player scores an average of 0.4 tries per match in which he plays.

\begin{enumerate}[label=(\alph*)]
\item Find the probability that he scores 2 or more tries in a match. [5 marks]
\end{enumerate}

The team's coach moves the player to a different position in the team believing he will then score more frequently. In the next five matches he scores 6 tries.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Stating your hypotheses clearly, test at the 5% level of significance whether or not there is evidence of an increase in the mean number of tries the player scores per match as a result of playing in a different position. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q4 [10]}}