| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Conditional probability or memoryless |
| Difficulty | Standard +0.3 This is a straightforward application of standard exponential distribution properties (mean, CDF, memoryless property) and Poisson process connection. All parts follow directly from formulas with minimal problem-solving required, though it covers multiple concepts across several marks making it slightly above average routine difficulty. |
| Spec | 5.02i 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 modelling5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= \frac{1}{\lambda} = 40\) hours | B1 | AO1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{time} < 12) = 1 - e^{-12 \times 0.025} = 1 - e^{-0.3} = 0.259\) | B1 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Exponential distribution has no memory | M1 | AO3.4 — Uses 'no memory' property |
| \(P(\text{time} > 30) = e^{-30 \times 0.025} = e^{-0.75} = 0.472\) | A1 | AO1.1b — Obtains probability |
| Answer | Marks | Guidance |
|---|---|---|
| 4 consecutive shifts gives \(4 \times 12 = 48\) hours | B1 | AO3.4 — States or uses new mean (or uses \(e^{-0.3}\)) |
| \(P(\text{time} > 48) = e^{-48 \times 0.025} = e^{-1.2}\) | M1 | AO1.1a — Finds probability \(P(\text{time} > 48)\) (or uses \((e^{-0.3})^4\)) |
| \(= 0.301\) | A1 | AO1.1b — Obtains correct probability |
| Answer | Marks | Guidance |
|---|---|---|
| Poisson identified | B1 | AO3.3 — States Poisson for model of situation given |
| \(\text{Po}(0.025)\) per hour | B1 | AO3.3 — States value for \(\lambda\) (= 0.025 per hour) for model |
| Answer | Marks | Guidance |
|---|---|---|
| For 60 hours of process: \(\lambda = 60 \times 0.025 = 1.5\) | M1F | AO3.4 — Uses 60 × 'their' \(\lambda\) for 'their' Po model, or uses exponential model to find \(P(\text{Time} > 60)\) |
| \(P(X = 0) = \frac{1.5^0 \times e^{-1.5}}{0!} = e^{-1.5} = 0.223\) | A1 | AO1.1b — Obtains correct probability using Poisson or exponential model |
# Question 8:
## Part (a):
Mean $= \frac{1}{\lambda} = 40$ hours | B1 | AO1.2
## Part (b):
$P(\text{time} < 12) = 1 - e^{-12 \times 0.025} = 1 - e^{-0.3} = 0.259$ | B1 | AO1.1b
## Part (c):
Exponential distribution has no memory | M1 | AO3.4 — Uses 'no memory' property
$P(\text{time} > 30) = e^{-30 \times 0.025} = e^{-0.75} = 0.472$ | A1 | AO1.1b — Obtains probability
## Part (d):
4 consecutive shifts gives $4 \times 12 = 48$ hours | B1 | AO3.4 — States or uses new mean (or uses $e^{-0.3}$)
$P(\text{time} > 48) = e^{-48 \times 0.025} = e^{-1.2}$ | M1 | AO1.1a — Finds probability $P(\text{time} > 48)$ (or uses $(e^{-0.3})^4$)
$= 0.301$ | A1 | AO1.1b — Obtains correct probability
## Part (e)(i):
Poisson identified | B1 | AO3.3 — States Poisson for model of situation given
$\text{Po}(0.025)$ per hour | B1 | AO3.3 — States value for $\lambda$ (= 0.025 per hour) for model
## Part (e)(ii):
For **60** hours of process: $\lambda = 60 \times 0.025 = 1.5$ | M1F | AO3.4 — Uses 60 × 'their' $\lambda$ for 'their' Po model, or uses exponential model to find $P(\text{Time} > 60)$
$P(X = 0) = \frac{1.5^0 \times e^{-1.5}}{0!} = e^{-1.5} = 0.223$ | A1 | AO1.1b — Obtains correct probability using Poisson or exponential model
---
**Total: 11 marks**
**Overall Total: 50 marks**8 The time in hours to failure of a component may be modelled by an exponential distribution with parameter $\lambda = 0.025$
In a manufacturing process, the machine involved uses one of these components continuously until it fails.
The component is then immediately replaced.\\
8
\begin{enumerate}[label=(\alph*)]
\item Write down the mean time to failure for a component.
8
\item Find the probability that a component will fail during a 12-hour shift.
8
\item A component has not failed for 30 hours. Find the probability that this component lasts for at least another 30 hours.\\[0pt]
[2 marks]
8
\item Find the probability that a component does not fail during 4 consecutive 12-hour shifts.\\[0pt]
[3 marks]\\
8
\item (i) State the distribution that can be used to model the number of components that fail during one hour of the manufacturing process.\\[0pt]
[2 marks]\\
8 (e) (ii) Hence, or otherwise, find the probability that no components fail during 5 consecutive 12-hour shifts.\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics Q8 [11]}}