Question 8 - A-Level Maths

AQA S2 2010 January — Question 8 18 marks

Exam Board	AQA
Module	S2 (Statistics 2)
Year	2010
Session	January
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Calculate and compare mean, median, mode
Difficulty	Standard +0.3 This is a standard S2 continuous probability distribution question requiring routine integration and application of standard formulas for expectation and variance. Parts (a)-(c) involve straightforward sketching and integration, part (d)(i) uses the variance formula Var(X) = E(X²) - [E(X)]², and parts (d)(ii)-(iii) apply linearity of expectation and variance properties—all textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration 5.03d E(g(X)): general expectation formula

8 The continuous random variable $X$ has probability density function given by $$\mathrm { f } ( x ) = \left\{ \begin{array} { c c } \frac { 1 } { 2 } \left( x ^ { 2 } + 1 \right) & 0 \leqslant x \leqslant 1 \\ ( x - 2 ) ^ { 2 } & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{array} \right.$$

Sketch the graph of f.
Calculate $\mathrm { P } ( X \leqslant 1 )$.
Show that $\mathrm { E } \left( X ^ { 2 } \right) = \frac { 4 } { 5 }$.
1. Given that $\mathrm { E } ( X ) = \frac { 19 } { 24 }$ and that $\operatorname { Var } ( X ) = \frac { 499 } { k }$, find the numerical value of $k$.
2. Find $\mathrm { E } \left( 5 X ^ { 2 } + 24 X - 3 \right)$.
3. Find $\operatorname { Var } ( 12 X - 5 )$.

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Question 8(a):

Answer	Marks	Guidance
Graph: curve from $(0, 0.5)$ to $(1, 1)$, then from $(1, 1)$ to $(2, 0)$	B3	B1 for axes; B1 for curve $(0, 0.5)$ to $(1, 1)$; B1 for curve $(1, 1)$ to $(2, 0)$

Question 8(b):

Answer	Marks	Guidance
$P(X \leq 1) = \int_0^1 \frac{1}{2}(x^2 + 1)\,dx = \left[\frac{x^3}{6} + \frac{x}{2}\right]_0^1 = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}$	M1, A1, A1	0.667

Question 8(c):

Answer	Marks	Guidance
$E(X^2) = \int_0^1 x^2 \times \frac{1}{2}(x^2+1)\,dx + \int_1^2 x^2(x-2)^2\,dx$	M1	both integrals seen
$= \left[\frac{x^5}{10} + \frac{x^3}{6}\right]_{x=0}^{x=1} + \left[\frac{x^5}{5} - x^4 + \frac{4x^3}{3}\right]_{x=1}^{x=2}$	A1A1
$= \left(\frac{1}{10} + \frac{1}{6}\right) + \left(\left[\frac{32}{5} - 16 + \frac{32}{3}\right] - \left[\frac{1}{5} - 1 + \frac{4}{3}\right]\right) = \frac{4}{5}$	m1, A1	dep(M1); AG

Question 8(d)(i):

Answer	Marks	Guidance
$E(X) = \frac{19}{24}$ and $k\text{Var}(X) = 499$; $\text{Var}(X) = E(X^2) - E^2(X) = \frac{4}{5} - \left(\frac{19}{24}\right)^2 = \frac{499}{2880}$ $(0.173)$; $\Rightarrow k = 2880$	M1, A1, A1	CAO

Question 8(d)(ii):

Answer	Marks	Guidance
$E(5X^2 + 24X - 3) = 5E(X^2) + 24E(X) - 3 = 5 \times \frac{4}{5} + 24 \times \frac{19}{24} - 3 = 20$	M1, A1	CAO

Question 8(d)(iii):

Answer	Marks	Guidance
$\text{Var}(12X - 5) = 144\text{Var}(X) = 144 \times \frac{499}{2880} = \frac{499}{20}$ or $(24.95)$	M1, A1	CAO (AWFW 24.9 to 25)

## Question 8(a):

| Graph: curve from $(0, 0.5)$ to $(1, 1)$, then from $(1, 1)$ to $(2, 0)$ | B3 | B1 for axes; B1 for curve $(0, 0.5)$ to $(1, 1)$; B1 for curve $(1, 1)$ to $(2, 0)$ |

## Question 8(b):

| $P(X \leq 1) = \int_0^1 \frac{1}{2}(x^2 + 1)\,dx = \left[\frac{x^3}{6} + \frac{x}{2}\right]_0^1 = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}$ | M1, A1, A1 | 0.667 |

## Question 8(c):

| $E(X^2) = \int_0^1 x^2 \times \frac{1}{2}(x^2+1)\,dx + \int_1^2 x^2(x-2)^2\,dx$ | M1 | both integrals seen |
| $= \left[\frac{x^5}{10} + \frac{x^3}{6}\right]_{x=0}^{x=1} + \left[\frac{x^5}{5} - x^4 + \frac{4x^3}{3}\right]_{x=1}^{x=2}$ | A1A1 | |
| $= \left(\frac{1}{10} + \frac{1}{6}\right) + \left(\left[\frac{32}{5} - 16 + \frac{32}{3}\right] - \left[\frac{1}{5} - 1 + \frac{4}{3}\right]\right) = \frac{4}{5}$ | m1, A1 | dep(M1); AG |

## Question 8(d)(i):

| $E(X) = \frac{19}{24}$ and $k\text{Var}(X) = 499$; $\text{Var}(X) = E(X^2) - E^2(X) = \frac{4}{5} - \left(\frac{19}{24}\right)^2 = \frac{499}{2880}$ $(0.173)$; $\Rightarrow k = 2880$ | M1, A1, A1 | CAO |

## Question 8(d)(ii):

| $E(5X^2 + 24X - 3) = 5E(X^2) + 24E(X) - 3 = 5 \times \frac{4}{5} + 24 \times \frac{19}{24} - 3 = 20$ | M1, A1 | CAO |

## Question 8(d)(iii):

| $\text{Var}(12X - 5) = 144\text{Var}(X) = 144 \times \frac{499}{2880} = \frac{499}{20}$ or $(24.95)$ | M1, A1 | CAO (AWFW 24.9 to 25) |

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8 The continuous random variable $X$ has probability density function given by

$$\mathrm { f } ( x ) = \left\{ \begin{array} { c c } 
\frac { 1 } { 2 } \left( x ^ { 2 } + 1 \right) & 0 \leqslant x \leqslant 1 \\
( x - 2 ) ^ { 2 } & 1 \leqslant x \leqslant 2 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of f.
\item Calculate $\mathrm { P } ( X \leqslant 1 )$.
\item Show that $\mathrm { E } \left( X ^ { 2 } \right) = \frac { 4 } { 5 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $\mathrm { E } ( X ) = \frac { 19 } { 24 }$ and that $\operatorname { Var } ( X ) = \frac { 499 } { k }$, find the numerical value of $k$.
\item Find $\mathrm { E } \left( 5 X ^ { 2 } + 24 X - 3 \right)$.
\item Find $\operatorname { Var } ( 12 X - 5 )$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA S2 2010 Q8 [18]}}

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Q1 5 Q2 5 Q3 7 Q4 10 Q5 10 Q6 10 Q7 10 Q8 18

Answer	Marks	Guidance
\(E(X^2) = \int_0^1 x^2 \times \frac{1}{2}(x^2+1)\,dx + \int_1^2 x^2(x-2)^2\,dx\)	M1	both integrals seen
\(= \left[\frac{x^5}{10} + \frac{x^3}{6}\right]_{x=0}^{x=1} + \left[\frac{x^5}{5} - x^4 + \frac{4x^3}{3}\right]_{x=1}^{x=2}\)	A1A1
\(= \left(\frac{1}{10} + \frac{1}{6}\right) + \left(\left[\frac{32}{5} - 16 + \frac{32}{3}\right] - \left[\frac{1}{5} - 1 + \frac{4}{3}\right]\right) = \frac{4}{5}\)	m1, A1	dep(M1); AG