| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Derive CDF from PDF |
| Difficulty | Moderate -0.8 Part (a) is a routine integration of a standard exponential PDF to derive the well-known CDF formula—a direct application of calculus with no problem-solving required. Part (b) is trivial substitution into the result. This is easier than average A-level content, being a standard bookwork question testing basic recall and technique. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(x) = \int_0^x \lambda e^{-\lambda t} \, dt\) | M1 (AO 1.1a) | Uses \(\int \lambda e^{-\lambda t} \, dt\); condone missing \(dt\) or using \(x\) for \(t\) |
| \(= \left[-e^{-\lambda t}\right]_0^x = 1 - e^{-\lambda x}\) | A1 (AO 1.1b) | Obtains correct integrated function; may be unsimplified |
| Completes argument by substituting limits and subtracting to show \(F(x) = 1 - e^{-\lambda x}\), or solving \(F(0) = 0\) to find constant | R1 (AO 2.1) | Condone missing \(dt\) or using \(x\) for \(t\); no other errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X > 1) = 1 - F(1) = e^{-2 \times 1}\) | M1 (AO 1.1a) | Obtains \(F(1) =\) AWRT \(0.865\) or selects correct integral; PI by correct final answer |
| \(= 0.135\) | A1 (AO 1.1b) | AWRT \(0.135\) |
## Question 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = \int_0^x \lambda e^{-\lambda t} \, dt$ | M1 (AO 1.1a) | Uses $\int \lambda e^{-\lambda t} \, dt$; condone missing $dt$ or using $x$ for $t$ |
| $= \left[-e^{-\lambda t}\right]_0^x = 1 - e^{-\lambda x}$ | A1 (AO 1.1b) | Obtains correct integrated function; may be unsimplified |
| Completes argument by substituting limits and subtracting to show $F(x) = 1 - e^{-\lambda x}$, or solving $F(0) = 0$ to find constant | R1 (AO 2.1) | Condone missing $dt$ or using $x$ for $t$; no other errors |
## Question 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X > 1) = 1 - F(1) = e^{-2 \times 1}$ | M1 (AO 1.1a) | Obtains $F(1) =$ **AWRT** $0.865$ or selects correct integral; PI by correct final answer |
| $= 0.135$ | A1 (AO 1.1b) | **AWRT** $0.135$ |3 The random variable $X$ has an exponential distribution with probability density function $\mathrm { f } ( x ) = \lambda \mathrm { e } ^ { - \lambda x }$ where $x \geq 0$
3
\begin{enumerate}[label=(\alph*)]
\item Show that the cumulative distribution function, for $x \geq 0$, is given by $\mathrm { F } ( x ) = 1 - \mathrm { e } ^ { - \lambda x }$\\[0pt]
[3 marks]\\
3
\item Given that $\lambda = 2$, find $\mathrm { P } ( X > 1 )$, giving your answer to three decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2022 Q3 [5]}}