Edexcel S3 — Question 8 20 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Marks20
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of a Poisson distribution
TypeComment on test validity or assumptions
DifficultyStandard +0.3 This is a standard S3 goodness-of-fit test question following a well-established template: estimate parameter, verify Poisson assumption (mean ≈ variance), perform chi-squared test with clear steps. While it requires multiple techniques and careful calculation (20 marks total), each component is routine for S3 students with no novel problem-solving required. The structure is predictable and similar to textbook exercises.
Spec5.02b Expectation and variance: discrete random variables5.06c Fit other distributions: discrete and continuous

A physicist believes that the number of particles emitted by a radioactive source with a long half-life can be modelled by a Poisson distribution. She records the number of particles emitted in 80 successive 5-minute periods and her results are shown in the table below.
No. of Particles012345 or more
No. of Intervals233214830
  1. Comment on the suitability of a Poisson distribution for this situation. [3 marks]
  2. Show that an unbiased estimate of the mean number of particles emitted in a 5-minute period is 1.2 and find an unbiased estimate of the variance. [5 marks]
  3. Explain how your answers to part (b) support the fitting of a Poisson distribution. [1 mark]
  4. Stating your hypotheses clearly and using a 5\% level of significance, test whether or not these data can be modelled by a Poisson distribution. [11 marks]

AnswerMarks
(a) e.g. particles are emitted singly, at random and at a constant rate (for near future given long half-life) so seems suitableB3
(b) \(\mu = \bar{x} = \frac{\Sigma f_i}{N} = \frac{96}{80} = 1.2\)M1 A1
\(\Sigma fx^2 = 32 + 56 + 72 + 48 = 208\)M1
\(\hat{\sigma}^2 = s^2 = \frac{80}{79}(\frac{208}{80} − 1.2^2) = 1.17\)M1 A1
(c) variance = mean as would be expected with a Poisson distributionB1
(d) \(H_0 :\) Po(1.2) is a suitable modelB1
\(H_1 :\) Po(1.2) is not a suitable model
\(P(0) = e^{−1.2} = 0.3012\)
\(P(1) = 1.2e^{−1.2} = 0.3614\)
\(P(2) = \frac{1.2^2e^{−1.2}}{2} = 0.2169\)
AnswerMarks
\(P(3) = \frac{1.2^3e^{−1.2}}{3×2} = 0.0867\)M1 A2
\(P(4) = \frac{1.2^4e^{−1.2}}{4×3×2} = 0.0260\)
× 80 to give exp. freqs then freq of ≥ 5 = (80 − sum of others)
AnswerMarks Guidance
\(\therefore\) exp. freqs are 24.10, 28.91, 17.35, 6.94, 2.08, 0.62M1 A1
combining groups ≥ 3M1
OE (O−E)
2324.10 −1.1
3228.91 3.09
1417.35 −3.35
119.64 1.36
\(\therefore \Sigma \frac{(O-E)^2}{E} = 1.219\)M1 A1
\(\nu = 4 − 2 = 2, \chi^2_{\text{crit}}(5\%) = 5.991\)M1
1.219 < 5.991 \(\therefore\) do not reject \(H_0\)
AnswerMarks Guidance
Po(1.2) is a suitable modelA1 (20)
Total(75)
**(a)** e.g. particles are emitted singly, at random and at a constant rate (for near future given long half-life) so seems suitable | B3 |

**(b)** $\mu = \bar{x} = \frac{\Sigma f_i}{N} = \frac{96}{80} = 1.2$ | M1 A1 |
$\Sigma fx^2 = 32 + 56 + 72 + 48 = 208$ | M1 |
$\hat{\sigma}^2 = s^2 = \frac{80}{79}(\frac{208}{80} − 1.2^2) = 1.17$ | M1 A1 |

**(c)** variance = mean as would be expected with a Poisson distribution | B1 |

**(d)** $H_0 :$ Po(1.2) is a suitable model | B1 |
$H_1 :$ Po(1.2) is not a suitable model

$P(0) = e^{−1.2} = 0.3012$
$P(1) = 1.2e^{−1.2} = 0.3614$
$P(2) = \frac{1.2^2e^{−1.2}}{2} = 0.2169$
$P(3) = \frac{1.2^3e^{−1.2}}{3×2} = 0.0867$ | M1 A2 |
$P(4) = \frac{1.2^4e^{−1.2}}{4×3×2} = 0.0260$

× 80 to give exp. freqs then freq of ≥ 5 = (80 − sum of others)
$\therefore$ exp. freqs are 24.10, 28.91, 17.35, 6.94, 2.08, 0.62 | M1 A1 |
combining groups ≥ 3 | M1 |

| O | E | (O−E) | $\frac{(O-E)^2}{E}$ |
|---|---|-------|-----|
| 23 | 24.10 | −1.1 | 0.0502 |
| 32 | 28.91 | 3.09 | 0.3303 |
| 14 | 17.35 | −3.35 | 0.6468 |
| 11 | 9.64 | 1.36 | 0.1919 |

$\therefore \Sigma \frac{(O-E)^2}{E} = 1.219$ | M1 A1 |
$\nu = 4 − 2 = 2, \chi^2_{\text{crit}}(5\%) = 5.991$ | M1 |
1.219 < 5.991 $\therefore$ do not reject $H_0$
Po(1.2) is a suitable model | A1 | (20)

**Total** | (75)
A physicist believes that the number of particles emitted by a radioactive source with a long half-life can be modelled by a Poisson distribution. She records the number of particles emitted in 80 successive 5-minute periods and her results are shown in the table below.

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
No. of Particles & 0 & 1 & 2 & 3 & 4 & 5 or more \\
\hline
No. of Intervals & 23 & 32 & 14 & 8 & 3 & 0 \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}[label=(\alph*)]
\item Comment on the suitability of a Poisson distribution for this situation. [3 marks]
\item Show that an unbiased estimate of the mean number of particles emitted in a 5-minute period is 1.2 and find an unbiased estimate of the variance. [5 marks]
\item Explain how your answers to part (b) support the fitting of a Poisson distribution. [1 mark]
\item Stating your hypotheses clearly and using a 5\% level of significance, test whether or not these data can be modelled by a Poisson distribution. [11 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S3  Q8 [20]}}