Challenging +1.2 This is a multi-step variance calculation requiring: (1) finding the linear function for 1<x≤3/2 using continuity and the integral condition, (2) computing E(X) and E(X²) with piecewise integration, and (3) applying Var(X)=E(X²)-[E(X)]². While it involves several techniques and careful algebra, it follows a standard template for piecewise PDF problems with no novel conceptual insight required. The 'show that' format provides a target to verify against, reducing difficulty slightly.
8 The continuous random variable \(X\) has probability density function \(\mathrm { f } ( x )\)
It is given that \(\mathrm { f } ( x ) = x ^ { 2 }\) for \(0 \leq x \leq 1\)
It is also given that \(\mathrm { f } ( x )\) is a linear function for \(1 < x \leq \frac { 3 } { 2 }\)
For all other values of \(x , \mathrm { f } ( x ) = 0\)
A sketch of the graph of \(y = \mathrm { f } ( x )\) is shown below.
\includegraphics[max width=\textwidth, alt={}, center]{c309e27b-5618-4f94-aecd-a55d8756ef03-12_821_1077_758_543}
Show that \(\operatorname { Var } ( X ) = 0.0864\) correct to three significant figures.
\includegraphics[max width=\textwidth, alt={}, center]{c309e27b-5618-4f94-aecd-a55d8756ef03-14_2491_1755_173_123}
Additional page, if required. number
Write the question numbers in the left-hand margin.
AO 1.1a. Uses the formula for \(\text{E}(g(X))\) to obtain a correct expression using their equation of the straight-line section for \(\text{E}(X)\) or \(\text{E}(X^2)\). PI. Condone missing brackets
AO 1.1a. Uses the formula for the variance to obtain a calculation to find \(\text{Var}(X)\)
\(\text{Var}(X) = 0.0864\) (3sf)
R1
AO 2.1. Completes a reasoned argument to obtain the given value of \(\text{Var}(X)\). A more accurate value needs to be seen before rounding, AWRT 0.08638
Question total: 8 marks
## Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \dfrac{\frac{5}{3}-1}{\frac{3}{2}-1} = \dfrac{4}{3}$, $f(x)-1 = \dfrac{4}{3}(x-1)$ for $1 < x \leq \dfrac{3}{2}$ | M1 | AO 3.1a. Uses one of the points shown on the graph and gradient $\frac{4}{3}$ to find an equation of the straight-line section of the pdf. **PI** |
| $f(x) = \dfrac{4}{3}x - \dfrac{1}{3}$ for $1 < x \leq \dfrac{3}{2}$ | A1 | AO 1.1b. Obtains a correct expression $\frac{4}{3}x - \frac{1}{3}$ for the straight-line section of the pdf |
| $\text{E}(X) = \displaystyle\int_0^1 x^3\,dx + \int_1^{\frac{3}{2}} x\!\left(\dfrac{4}{3}x - \dfrac{1}{3}\right)dx = \dfrac{1}{4} + \dfrac{61}{72} = \dfrac{79}{72}$ | M1 | AO 1.1a. Uses the formula for $\text{E}(g(X))$ to obtain a correct expression using their equation of the straight-line section for $\text{E}(X)$ or $\text{E}(X^2)$. **PI**. Condone missing brackets |
| $\text{E}(X) = \dfrac{79}{72}$ and $\text{E}(X^2) = \displaystyle\int_0^1 x^4\,dx + \int_1^{\frac{3}{2}} x^2\!\left(\dfrac{4}{3}x - \dfrac{1}{3}\right)dx = \dfrac{1}{5} + \dfrac{157}{144} = \dfrac{929}{720}$ | A1F | AO 1.1b. Obtain correct expressions for $\text{E}(X)$ and $\text{E}(X^2)$. **PI**. **FT** their equation of the straight-line section |
| $\text{E}(X) = \dfrac{79}{72}$, **AWRT** 1.10 | A1 | AO 1.1b. Obtains the correct value of $\text{E}(X)$. **PI** by correct calculation substituted into calculation to find $\text{Var}(X)$ |
| $\text{E}(X^2) = \dfrac{929}{720}$, **AWRT** 1.29 | A1 | AO 1.1b. Obtains the correct value of $\text{E}(X^2)$. **PI** by correct calculation substituted into calculation to find $\text{Var}(X)$ |
| $\text{Var}(X) = \text{E}(X^2) - (\text{E}(X))^2 = \dfrac{929}{720} - \left(\dfrac{79}{72}\right)^2 = 0.08638 = 0.0864$ (3sf) | M1 | AO 1.1a. Uses the formula for the variance to obtain a calculation to find $\text{Var}(X)$ |
| $\text{Var}(X) = 0.0864$ (3sf) | R1 | AO 2.1. Completes a reasoned argument to obtain the given value of $\text{Var}(X)$. A more accurate value needs to be seen before rounding, **AWRT** 0.08638 |
**Question total: 8 marks**
8 The continuous random variable $X$ has probability density function $\mathrm { f } ( x )$
It is given that $\mathrm { f } ( x ) = x ^ { 2 }$ for $0 \leq x \leq 1$\\
It is also given that $\mathrm { f } ( x )$ is a linear function for $1 < x \leq \frac { 3 } { 2 }$\\
For all other values of $x , \mathrm { f } ( x ) = 0$
A sketch of the graph of $y = \mathrm { f } ( x )$ is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{c309e27b-5618-4f94-aecd-a55d8756ef03-12_821_1077_758_543}
Show that $\operatorname { Var } ( X ) = 0.0864$ correct to three significant figures.\\
\includegraphics[max width=\textwidth, alt={}, center]{c309e27b-5618-4f94-aecd-a55d8756ef03-14_2491_1755_173_123}
Additional page, if required. number
Write the question numbers in the left-hand margin.
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2023 Q8 [8]}}