| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Calculate mean or variance |
| Difficulty | Standard +0.3 This is a standard proof question requiring integration by parts for both parts, which is a routine technique in Further Maths Statistics. While it requires careful algebraic manipulation and knowledge of the exponential distribution's pdf, these are textbook derivations that students are expected to memorize or reproduce mechanically. The question is slightly above average difficulty only because it's a proof rather than calculation, but it's a fundamental result with well-established methods. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \lim_{k\to\infty} \int_0^k \lambda x e^{-\lambda x}\ dx\) | M1 (AO1.1a) | Condone use of \(\int_0^\infty \lambda x e^{-\lambda x}\ dx\); condone use of \(-\infty\) instead of 0 or missing limits |
| \(= \lim_{k\to\infty}\left[-xe^{-\lambda x}\right]_0^k + \lim_{k\to\infty}\int_0^k e^{-\lambda x}\ dx\) | A1F (AO1.1b) | Uses integration by parts; FT integration of \(\lambda xe^{\lambda x}\) |
| \(= \lim_{k\to\infty}\left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^k\) | M1 (AO1.1a) | Integrates \(\int e^{-\lambda x}\ dx\) |
| \(= 0 - \left(-\frac{1}{\lambda}\right) = \frac{1}{\lambda}\) | R1 (AO2.1) | Completes rigorous argument; condone absence of limiting process throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = \lim_{k\to\infty} \int_0^k \lambda x^2 e^{-\lambda x}\, dx\) | M1 (1.1a) | Condone use of \(\int_0^\infty \lambda x^2 e^{-\lambda x}\, dx\); condone use of \(-\infty\) instead of 0 or missing limits; condone missing \(dx\); condone sign error on power |
| \(= \lim_{k\to\infty}\left[-x^2 e^{-\lambda x}\right]_0^k + \lim_{k\to\infty}\int_0^k 2xe^{-\lambda x}\, dx = \lim_{k\to\infty}\int_0^k 2xe^{-\lambda x}\, dx\) | A1F (1.1b) | Uses integration by parts once to simplify integral; condone missing \(dx\); FT integration of \(\lambda x^2 e^{\lambda x}\) |
| \(= \lim_{k\to\infty}\left[-\frac{2}{\lambda}xe^{-\lambda x}\right]_0^k + \lim_{k\to\infty}\int_0^k \frac{2}{\lambda}e^{-\lambda x}\, dx = \lim_{k\to\infty}\int_0^k \frac{2}{\lambda}e^{-\lambda x}\, dx = \lim_{k\to\infty}\left[-\frac{2}{\lambda^2}e^{-\lambda x}\right]_0^k\) | A1F (1.1b) | Uses integration by parts twice to simplify integral; condone missing \(dx\); FT integration of \(\lambda x^2 e^{\lambda x}\) |
| \(= 0 - \left(-\frac{2}{\lambda^2}\right) = \frac{2}{\lambda^2}\) | M1 (1.1a) | Integrates \(\int_0^\infty \frac{2}{\lambda}e^{-\lambda x}\, dx\) |
| \(\text{Var}(X) = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2\) | A1 (1.1b) | Obtains correct expression for \(E(X^2)\) |
| \(= \frac{1}{\lambda^2}\) | M1 (1.1a) | Substitutes into formula \(\text{Var}(X) = E(X^2) - (E(X))^2\) |
| Completes rigorous argument to find correct expression for \(\text{Var}(X)\) | R1 (2.1) | Condone absence of limiting process throughout |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \lim_{k\to\infty} \int_0^k \lambda x e^{-\lambda x}\ dx$ | M1 (AO1.1a) | Condone use of $\int_0^\infty \lambda x e^{-\lambda x}\ dx$; condone use of $-\infty$ instead of 0 or missing limits |
| $= \lim_{k\to\infty}\left[-xe^{-\lambda x}\right]_0^k + \lim_{k\to\infty}\int_0^k e^{-\lambda x}\ dx$ | A1F (AO1.1b) | Uses integration by parts; FT integration of $\lambda xe^{\lambda x}$ |
| $= \lim_{k\to\infty}\left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^k$ | M1 (AO1.1a) | Integrates $\int e^{-\lambda x}\ dx$ |
| $= 0 - \left(-\frac{1}{\lambda}\right) = \frac{1}{\lambda}$ | R1 (AO2.1) | Completes rigorous argument; condone absence of limiting process throughout |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = \lim_{k\to\infty} \int_0^k \lambda x^2 e^{-\lambda x}\, dx$ | M1 (1.1a) | Condone use of $\int_0^\infty \lambda x^2 e^{-\lambda x}\, dx$; condone use of $-\infty$ instead of 0 or missing limits; condone missing $dx$; condone sign error on power |
| $= \lim_{k\to\infty}\left[-x^2 e^{-\lambda x}\right]_0^k + \lim_{k\to\infty}\int_0^k 2xe^{-\lambda x}\, dx = \lim_{k\to\infty}\int_0^k 2xe^{-\lambda x}\, dx$ | A1F (1.1b) | Uses integration by parts once to simplify integral; condone missing $dx$; FT integration of $\lambda x^2 e^{\lambda x}$ |
| $= \lim_{k\to\infty}\left[-\frac{2}{\lambda}xe^{-\lambda x}\right]_0^k + \lim_{k\to\infty}\int_0^k \frac{2}{\lambda}e^{-\lambda x}\, dx = \lim_{k\to\infty}\int_0^k \frac{2}{\lambda}e^{-\lambda x}\, dx = \lim_{k\to\infty}\left[-\frac{2}{\lambda^2}e^{-\lambda x}\right]_0^k$ | A1F (1.1b) | Uses integration by parts twice to simplify integral; condone missing $dx$; FT integration of $\lambda x^2 e^{\lambda x}$ |
| $= 0 - \left(-\frac{2}{\lambda^2}\right) = \frac{2}{\lambda^2}$ | M1 (1.1a) | Integrates $\int_0^\infty \frac{2}{\lambda}e^{-\lambda x}\, dx$ |
| $\text{Var}(X) = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2$ | A1 (1.1b) | Obtains correct expression for $E(X^2)$ |
| $= \frac{1}{\lambda^2}$ | M1 (1.1a) | Substitutes into formula $\text{Var}(X) = E(X^2) - (E(X))^2$ |
| Completes rigorous argument to find correct expression for $\text{Var}(X)$ | R1 (2.1) | Condone absence of limiting process throughout |
---7 The random variable $X$ has an exponential distribution with parameter $\lambda$
7
\begin{enumerate}[label=(\alph*)]
\item Prove that $\mathrm { E } ( X ) = \frac { 1 } { \lambda }$\\
7
\item Prove that $\operatorname { Var } ( X ) = \frac { 1 } { \lambda ^ { 2 } }$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2021 Q7 [11]}}