AQA Further Paper 3 Statistics 2024 June — Question 5 5 marks

Exam BoardAQA
ModuleFurther Paper 3 Statistics (Further Paper 3 Statistics)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeFind expectation E(X)
DifficultyStandard +0.3 This is a straightforward expectation calculation requiring integration by parts of xe^(x/3), which is a standard technique taught in A-level Further Maths. The bounds are clean (0 to ln27) and simplify nicely. While it requires careful algebraic manipulation, it follows a routine procedure with no conceptual difficulty or novel insight required.
Spec5.02b Expectation and variance: discrete random variables

5 The continuous random variable \(X\) has probability density function $$f ( x ) = \begin{cases} \frac { 1 } { 6 } e ^ { \frac { x } { 3 } } & 0 \leq x \leq \ln 27 \\ 0 & \text { otherwise } \end{cases}$$ Show that the mean of \(X\) is \(\frac { 3 } { 2 } ( \ln 27 - 2 )\)

Question 5:
Setting up integral
AnswerMarks Guidance
\(\displaystyle\int_0^{\ln 27} \frac{x}{6} e^{\frac{1}{3}x}\, dx\)M1 Uses \(\displaystyle\int \frac{x}{6} e^{\frac{1}{3}x}\, dx\) with any or no limits; condone missing \(dx\)
Integration by parts
AnswerMarks Guidance
\(\left[kx e^{\frac{1}{3}x}\right] - \displaystyle\int k e^{\frac{1}{3}x}\, dx\)M1 Correct method for integration by parts; condone missing \(dx\)
Integrated function
AnswerMarks Guidance
\(\dfrac{x}{2}e^{\frac{1}{3}x} - \dfrac{3}{2}e^{\frac{1}{3}x}\)A1 OE; components may appear on different lines
Substituting limits
AnswerMarks Guidance
Substitutes limits of \(0\) and \(\ln 27\) and subtracts correctlyM1 Integral must have at least two terms; PI by sight of \(\mathbf{AWRT}\ 1.94\)
Final answer
AnswerMarks Guidance
\(= \dfrac{3}{2}(\ln 27 - 2)\)R1 Completes reasoned argument to obtain \(\dfrac{3}{2}(\ln 27 - 2)\)
## Question 5:

**Setting up integral**
$\displaystyle\int_0^{\ln 27} \frac{x}{6} e^{\frac{1}{3}x}\, dx$ | M1 | Uses $\displaystyle\int \frac{x}{6} e^{\frac{1}{3}x}\, dx$ with any or no limits; condone missing $dx$

**Integration by parts**
$\left[kx e^{\frac{1}{3}x}\right] - \displaystyle\int k e^{\frac{1}{3}x}\, dx$ | M1 | Correct method for integration by parts; condone missing $dx$

**Integrated function**
$\dfrac{x}{2}e^{\frac{1}{3}x} - \dfrac{3}{2}e^{\frac{1}{3}x}$ | A1 | OE; components may appear on different lines

**Substituting limits**
Substitutes limits of $0$ and $\ln 27$ and subtracts correctly | M1 | Integral must have at least two terms; PI by sight of $\mathbf{AWRT}\ 1.94$

**Final answer**
$= \dfrac{3}{2}(\ln 27 - 2)$ | R1 | Completes reasoned argument to obtain $\dfrac{3}{2}(\ln 27 - 2)$

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5 The continuous random variable $X$ has probability density function

$$f ( x ) = \begin{cases} \frac { 1 } { 6 } e ^ { \frac { x } { 3 } } & 0 \leq x \leq \ln 27 \\ 0 & \text { otherwise } \end{cases}$$

Show that the mean of $X$ is $\frac { 3 } { 2 } ( \ln 27 - 2 )$\\

\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2024 Q5 [5]}}