Question 6 - A-Level Maths

Edexcel S2 2016 June — Question 6 11 marks

Exam Board	Edexcel
Module	S2 (Statistics 2)
Year	2016
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	PDF with multiple constants
Difficulty	Standard +0.3 This is a standard S2 PDF question requiring routine calculus techniques: finding the mode by differentiation, using the total probability condition to find constants, computing a CDF value by integration, and interpreting quartiles. All steps are textbook procedures with no novel insight required, making it slightly easier than average.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03e Find cdf: by integration 5.03f Relate pdf-cdf: medians and percentiles

6. A continuous random variable $X$ has probability density function $$f ( x ) = \begin{cases} a x - b x ^ { 2 } & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$ Given that the mode is 1

show that $a = 2 b$
Find the value of $a$ and the value of $b$
Calculate F(1.5)
State whether the upper quartile of $X$ is greater than 1.5, equal to 1.5, or less than 1.5 Give a reason for your answer.

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Question 6:

Part (a):

Answer	Marks	Guidance
$\frac{d}{dx}(ax - bx^2) = a - 2bx$	M1	Differentiating $f(x)$, at least one $x^n \to x^{n-1}$, must lead to function of $x$
$a - 2b(1) = 0 \Rightarrow a = 2b$	A1cso	Fully correct solution with no errors seen. Beware: Use of $f(2)=0$ scores M0A0.

Part (b):

Answer	Marks	Guidance
$\int_0^2 (ax - bx^2)\,dx = 1$	M1	Attempt to integrate and equate to 1, at least one $x^n \to x^{n+1}$, ignore limits
$\left[\frac{ax^2}{2} - \frac{bx^3}{3}\right]_0^2 = 1$	A1	Correct integration (in terms of $a$ or $b$ or both) and sight of correct limits
$\frac{(2b)(2^2)}{2} - \frac{b(2^3)}{3} = 1 \quad a = \frac{3}{2},\quad b = \frac{3}{4}$	dM1 $\underline{A1}\ \underline{A1}$	$2^{nd}$ dM1 for use of correct limits (at least $x=2$ must be seen) and substituting $a=2b$ to obtain equation in 1 variable (dependent on previous M1). NB sight of $2a - \frac{8}{3}b = 1$ scores first M1A1.

Part (c):

Answer	Marks	Guidance
$\int_0^{1.5} f(x)\,dx = \left[\frac{ax^2}{2} - \frac{bx^3}{3}\right]_0^{1.5}$	M1	For use of $F(1.5)$ or $\int_0^{1.5} f(x)\,dx$, at least one $x^n \to x^{n+1}$, with limits and ft their $a$ and $b$
$= \frac{\frac{3}{2}(1.5)^2}{2} - \frac{(\frac{3}{4})(1.5)^3}{3} = \frac{27}{32}$ or awrt 0.844	A1

Part (d):

Answer	Marks	Guidance
$F(1.5) > 0.75$	M1	Correct comparison of their $F(1.5)$ with 0.75
Therefore the upper quartile of $X$ is less than 1.5	A1ft	ft for correct conclusion provided $0.5 < F(1.5) < 1$. Find $Q_3$: M1 if attempt $F(x) = 0.75$ and get $Q_3 =$ awrt 1.35 (calc 1.347296...) and A1 for conclusion.

## Question 6:

### Part (a):
$\frac{d}{dx}(ax - bx^2) = a - 2bx$ | M1 | Differentiating $f(x)$, at least one $x^n \to x^{n-1}$, must lead to function of $x$

$a - 2b(1) = 0 \Rightarrow a = 2b$ | A1cso | Fully correct solution with no errors seen. **Beware:** Use of $f(2)=0$ scores M0A0.

### Part (b):
$\int_0^2 (ax - bx^2)\,dx = 1$ | M1 | Attempt to integrate and equate to 1, at least one $x^n \to x^{n+1}$, ignore limits

$\left[\frac{ax^2}{2} - \frac{bx^3}{3}\right]_0^2 = 1$ | A1 | Correct integration (in terms of $a$ or $b$ or both) and sight of correct limits

$\frac{(2b)(2^2)}{2} - \frac{b(2^3)}{3} = 1 \quad a = \frac{3}{2},\quad b = \frac{3}{4}$ | dM1 $\underline{A1}\ \underline{A1}$ | $2^{nd}$ dM1 for use of correct limits (at least $x=2$ must be seen) and substituting $a=2b$ to obtain equation in 1 variable (dependent on previous M1). NB sight of $2a - \frac{8}{3}b = 1$ scores first M1A1.

### Part (c):
$\int_0^{1.5} f(x)\,dx = \left[\frac{ax^2}{2} - \frac{bx^3}{3}\right]_0^{1.5}$ | M1 | For use of $F(1.5)$ or $\int_0^{1.5} f(x)\,dx$, at least one $x^n \to x^{n+1}$, with limits and ft their $a$ and $b$

$= \frac{\frac{3}{2}(1.5)^2}{2} - \frac{(\frac{3}{4})(1.5)^3}{3} = \frac{27}{32}$ or awrt **0.844** | A1 |

### Part (d):
$F(1.5) > 0.75$ | M1 | Correct comparison of their $F(1.5)$ with 0.75

Therefore the upper quartile of $X$ is less than 1.5 | A1ft | ft for correct conclusion provided $0.5 < F(1.5) < 1$. **Find $Q_3$:** M1 if attempt $F(x) = 0.75$ **and** get $Q_3 =$ awrt 1.35 (calc 1.347296...) and A1 for conclusion.

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6. A continuous random variable $X$ has probability density function

$$f ( x ) = \begin{cases} a x - b x ^ { 2 } & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$

Given that the mode is 1
\begin{enumerate}[label=(\alph*)]
\item show that $a = 2 b$
\item Find the value of $a$ and the value of $b$
\item Calculate F(1.5)
\item State whether the upper quartile of $X$ is greater than 1.5, equal to 1.5, or less than 1.5 Give a reason for your answer.\\

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2016 Q6 [11]}}

This paper (7 questions)

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Q1 11 Q2 10 Q3 6 Q4 13 Q5 9 Q6 11 Q7 15

Answer	Marks	Guidance
\(\int_0^2 (ax - bx^2)\,dx = 1\)	M1	Attempt to integrate and equate to 1, at least one \(x^n \to x^{n+1}\), ignore limits
\(\left[\frac{ax^2}{2} - \frac{bx^3}{3}\right]_0^2 = 1\)	A1	Correct integration (in terms of \(a\) or \(b\) or both) and sight of correct limits
\(\frac{(2b)(2^2)}{2} - \frac{b(2^3)}{3} = 1 \quad a = \frac{3}{2},\quad b = \frac{3}{4}\)	dM1 \(\underline{A1}\ \underline{A1}\)	\(2^{nd}\) dM1 for use of correct limits (at least \(x=2\) must be seen) and substituting \(a=2b\) to obtain equation in 1 variable (dependent on previous M1). NB sight of \(2a - \frac{8}{3}b = 1\) scores first M1A1.

Answer	Marks	Guidance
\(\frac{d}{dx}(ax - bx^2) = a - 2bx\)	M1	Differentiating \(f(x)\), at least one \(x^n \to x^{n-1}\), must lead to function of \(x\)
\(a - 2b(1) = 0 \Rightarrow a = 2b\)	A1cso	Fully correct solution with no errors seen. Beware: Use of \(f(2)=0\) scores M0A0.

Answer	Marks	Guidance
\(\int_0^{1.5} f(x)\,dx = \left[\frac{ax^2}{2} - \frac{bx^3}{3}\right]_0^{1.5}\)	M1	For use of \(F(1.5)\) or \(\int_0^{1.5} f(x)\,dx\), at least one \(x^n \to x^{n+1}\), with limits and ft their \(a\) and \(b\)
\(= \frac{\frac{3}{2}(1.5)^2}{2} - \frac{(\frac{3}{4})(1.5)^3}{3} = \frac{27}{32}\) or awrt 0.844	A1

Answer	Marks	Guidance
\(F(1.5) > 0.75\)	M1	Correct comparison of their \(F(1.5)\) with 0.75
Therefore the upper quartile of \(X\) is less than 1.5	A1ft	ft for correct conclusion provided \(0.5 < F(1.5) < 1\). Find \(Q_3\): M1 if attempt \(F(x) = 0.75\) and get \(Q_3 =\) awrt 1.35 (calc 1.347296...) and A1 for conclusion.