| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2016 |
| Session | June |
| Marks | 22 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Proving Poisson properties from first principles |
| Difficulty | Standard +0.3 This is a standard S3 question covering core Poisson distribution material. Part (a) is a bookwork proof using standard series manipulation (E(X) and Var(X) for Poisson). Parts (b) and (c) involve routine applications: direct Poisson probability calculation, normal approximation to Poisson with continuity correction, and constructing a confidence interval for λ. All techniques are standard textbook exercises with no novel problem-solving required, making this slightly easier than average for A-level. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04b Linear combinations: of normal distributions5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \sum_{x=0}^{\infty} x \cdot \frac{e^{-\lambda}\lambda^x}{x!} =\) | M1 | Used; ignore limits until A1 |
| \(\lambda \sum \frac{e^{-\lambda}\lambda^{x-1}}{(x-1)!} =\) | M1 | Factor of \(\lambda\) plus x! to (x – 1)! |
| with \(y = x - 1\) | ||
| \(\lambda \sum_{y=0}^{\infty} \frac{e^{-\lambda}\lambda^y}{y!} = \lambda \times 1 = \lambda\) | A1 | Fully complete and correct derivation AG |
| 3 | ||
| \(E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \cdot \frac{e^{-\lambda}\lambda^x}{x!} =\) | M1 | Used; ignore limits until A1 |
| \(\lambda^2 \sum \frac{e^{-\lambda}\lambda^{x-2}}{(x-2)!} = \lambda^2\) | A1 | Factor of \(\lambda^2\) plus x! to (x – 2)! and fully complete and correct derivation |
| 2 | ||
| \(\text{Var}(X) = E(X^2) - (E(X))^2 =\) | M1 | Used |
| \(E(X(X-1)) + \lambda - \lambda^2 = \lambda\) | A1 | Fully complete and correct derivation |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Po(0.75) | B1 | PI |
| \(P(0 \text{ faults}) = e^{-0.75} = \mathbf{0.472}\) | B1 | AWRT (0.47237) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Po(37.5) \(\Rightarrow N(37.5, 37.5)\) | B1 | Normal with mean = variance = 37.5 in (A) or (B) |
| \(P(F < 30) = P\left(Z < \frac{29.5 - 37.5}{\sqrt{37.5}}\right)\) | M1 | Standardising (29.5 or 30 or 30.5) with C's mean = variance |
| \(= P(Z < -1.30639) = 1 - P(Z < 1.30639)\) | ml | Area change; can be implied by any final answer \(< 0.5\) |
| \(= \mathbf{0.095 \text{ to } 0.097}\) | A1 | AWFW (0.09571) |
| 4 | ||
| \(P(35 \leq F \leq 45) =\) | M1 | Area difference |
| \(P(F \leq 45.5 \text{ or } 45) - P(F \leq 34.5 \text{ or } 35) =\) | ||
| \(P(Z < 1.31) - P(Z < -0.49)\) | A1 | Both AWRT (1.30639 & 0.48990) |
| \(= \mathbf{0.591 \text{ to } 0.597}\) | A1 | AWFW (0.59219) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Use of Poisson: (A) 0.092 (AWRT) \(\Rightarrow\) B2 (B) 0.582 (AWRT) \(\Rightarrow\) B1 | (max of 3 marks) | |
| 7 | ||
| Total for (a) & (b) | 16 |
| Answer | Marks | Guidance |
|---|---|---|
| \(98\% \Rightarrow z = \mathbf{2.32 \text{ to } 2.33}\) | B1 | AWFW (2.3263) |
| CI: | M1 | \(\lambda \pm z\sqrt{a}\) |
| \(\begin{pmatrix} 49 \\ 4.9 \\ 0.98 \\ 0.098 \end{pmatrix} \pm \begin{pmatrix} 2.32 \text{ to } 2.33 \\ 2.05 \text{ to } 2.06 \end{pmatrix} \begin{pmatrix} \sqrt{49} = 7 \\ \sqrt{4.9/10} = 0.7 \\ \sqrt{0.98/50} = 0.14 \\ \sqrt{0.098/500} = 0.014 \end{pmatrix}\) | A1 | Any correct value for \(\lambda\) |
| A1 | Correct expression for a given \(\lambda\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.098 \pm (0.032 \text{ to } 0.034) = (0.064 \text{ to } 0.066, 0.130 \text{ to } 0.132)\) | ||
| Dividing by 500, 50, 10 or 1 as appropriate | B1 | CAO |
| ie \(\mathbf{0.098 \pm (0.032 \text{ to } 0.034)}\) | A1 | CAO ± AWFW (0.03257) |
| or \(\mathbf{(0.064 \text{ to } 0.066, 0.130 \text{ to } 0.132)}\) | AWFW | |
| 6 | ||
| Total | 22 |
## Part (a)
$E(X) = \sum_{x=0}^{\infty} x \cdot \frac{e^{-\lambda}\lambda^x}{x!} =$ | M1 | Used; ignore limits until A1
$\lambda \sum \frac{e^{-\lambda}\lambda^{x-1}}{(x-1)!} =$ | M1 | Factor of $\lambda$ plus x! to (x – 1)!
with $y = x - 1$ | |
$\lambda \sum_{y=0}^{\infty} \frac{e^{-\lambda}\lambda^y}{y!} = \lambda \times 1 = \lambda$ | A1 | Fully complete and correct derivation AG
| | **3** |
$E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \cdot \frac{e^{-\lambda}\lambda^x}{x!} =$ | M1 | Used; ignore limits until A1
$\lambda^2 \sum \frac{e^{-\lambda}\lambda^{x-2}}{(x-2)!} = \lambda^2$ | A1 | Factor of $\lambda^2$ plus x! to (x – 2)! and fully complete and correct derivation
| | **2** |
$\text{Var}(X) = E(X^2) - (E(X))^2 =$ | M1 | Used
$E(X(X-1)) + \lambda - \lambda^2 = \lambda$ | A1 | Fully complete and correct derivation
| | **2** |
## Part (b)(i)
Po(0.75) | B1 | PI
$P(0 \text{ faults}) = e^{-0.75} = \mathbf{0.472}$ | B1 | AWRT (0.47237)
| | **2** |
## Part (b)(ii)
Po(37.5) $\Rightarrow N(37.5, 37.5)$ | B1 | Normal with mean = variance = 37.5 in (A) or (B)
$P(F < 30) = P\left(Z < \frac{29.5 - 37.5}{\sqrt{37.5}}\right)$ | M1 | Standardising (29.5 or 30 or 30.5) with C's mean = variance
$= P(Z < -1.30639) = 1 - P(Z < 1.30639)$ | ml | Area change; can be implied by any final answer $< 0.5$
$= \mathbf{0.095 \text{ to } 0.097}$ | A1 | AWFW (0.09571)
| | **4** |
$P(35 \leq F \leq 45) =$ | M1 | Area difference
$P(F \leq 45.5 \text{ or } 45) - P(F \leq 34.5 \text{ or } 35) =$ | |
$P(Z < 1.31) - P(Z < -0.49)$ | A1 | Both AWRT (1.30639 & 0.48990)
$= \mathbf{0.591 \text{ to } 0.597}$ | A1 | AWFW (0.59219)
| | **3** |
## Special Case
Use of Poisson: (A) 0.092 (AWRT) $\Rightarrow$ B2 (B) 0.582 (AWRT) $\Rightarrow$ B1 | | (max of 3 marks)
| | **7** |
**Total for (a) & (b)** | | **16** |
## Part (c)
$98\% \Rightarrow z = \mathbf{2.32 \text{ to } 2.33}$ | B1 | AWFW (2.3263)
CI: | M1 | $\lambda \pm z\sqrt{a}$
$\begin{pmatrix} 49 \\ 4.9 \\ 0.98 \\ 0.098 \end{pmatrix} \pm \begin{pmatrix} 2.32 \text{ to } 2.33 \\ 2.05 \text{ to } 2.06 \end{pmatrix} \begin{pmatrix} \sqrt{49} = 7 \\ \sqrt{4.9/10} = 0.7 \\ \sqrt{0.98/50} = 0.14 \\ \sqrt{0.098/500} = 0.014 \end{pmatrix}$ | A1 | Any correct value for $\lambda$
| A1 | Correct expression for a given $\lambda$
or $49 \pm (16.2 \text{ to } 16.4) = (32.6 \text{ to } 32.8, 65.2 \text{ to } 65.4)$
$4.9 \pm (1.62 \text{ to } 1.64) = (3.26 \text{ to } 3.28, 6.52 \text{ to } 6.54)$
$0.98 \pm (0.32 \text{ to } 0.34) = (0.64 \text{ to } 0.66, 1.30 \text{ to } 1.32)$
$0.098 \pm (0.032 \text{ to } 0.034) = (0.064 \text{ to } 0.066, 0.130 \text{ to } 0.132)$ | |
Dividing by 500, 50, 10 or 1 as appropriate | B1 | CAO
ie $\mathbf{0.098 \pm (0.032 \text{ to } 0.034)}$ | A1 | CAO ± AWFW (0.03257)
or $\mathbf{(0.064 \text{ to } 0.066, 0.130 \text{ to } 0.132)}$ | | AWFW
| | **6** |
**Total** | | **22** |\begin{enumerate}[label=(\alph*)]
\item The discrete random variable $X$ has probability distribution given by
$$\mathrm{P}(X = x) = \begin{cases}
\frac{e^{-\lambda}\lambda^x}{x!} & x = 0, 1, 2, \ldots \\
0 & \text{otherwise}
\end{cases}$$
Show that $\mathrm{E}(X) = \lambda$ and that $\mathrm{Var}(X) = \lambda$. [7 marks]
\item In light-weight chain, faults occur randomly and independently, and at a constant average rate of 0.075 per metre.
\begin{enumerate}[label=(\roman*)]
\item Calculate the probability that there are no faults in a 10-metre length of this chain. [2 marks]
\item Use a distributional approximation to estimate the probability that, in a 500-metre reel of light-weight chain, there are:
\begin{enumerate}[label=(\textbf{\Alph*})]
\item fewer than 30 faults;
\item at least 35 faults but at most 45 faults.
\end{enumerate}
[7 marks]
\end{enumerate}
\item As part of an investigation into the quality of a new design of medium-weight chain, a sample of fifty 10-metre lengths was selected.
Subsequent analysis revealed a total of 49 faults.
Assuming that faults occur randomly and independently, and at a constant average rate, construct an approximate 98\% confidence interval for the average number of faults per metre. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2016 Q6 [22]}}