Question 3 - A-Level Maths

Edexcel S2 2017 June — Question 3 12 marks

Exam Board	Edexcel
Module	S2 (Statistics 2)
Year	2017
Session	June
Marks	12
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 integration to find constants and the CDF. Part (a) uses the median property (area = 0.5), parts (b) and (c) involve routine integration and applying continuity/normalization conditions. While multi-part, each step follows 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

3. The random variable $X$ has probability density function given by $$f ( x ) = \begin{cases} a x + b & 1 \leqslant x < 4 \\ \frac { 3 } { 2 } - \frac { 1 } { 4 } x & 4 \leqslant x \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$ as shown in Figure 1, where $a$ and $b$ are constants. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{1a1534ea-4c62-4945-850a-9460ea87fd64-08_634_1132_694_397} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure}

Show that the median of $X$ is 4
Find the value of $a$ and the value of $b$
Specify fully the cumulative distribution function of $X$

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) Using area of triangle: $P(X > 4) = \frac{(6-4) \times f(4)}{2} = \frac{1}{2}$ [so median of $X$ is 4] OR Using integration: $\int_4^6 \left(\frac{1}{3} - \frac{1}{4}x\right) dx = \left[\frac{1}{3}x - \frac{1}{8}x^2\right]_4 = \frac{1}{2}$ [so median of $X$ is 4]	M1, A1 cso	M1 for a correct expression for area above 4 using triangle or correct integral & attempt to integrate. A1 cso for $\frac{1}{2}$ or $x = 4$ with no errors. NB $f(4) = 1.5 - 1 = 0.5$ or $1.5 - 0.25x = 0.5$ is M0A0
(b) Area of triangle from 1 < $x$ < 4 = 0.5: $\frac{(4-1) \times (4a + b)}{2} = 0.5 \to 12a + 3b = 1$ OR $\int_1^4 (ax + b) dx = 0.5 \to 15a + 6b = 1$	M1, M1	1st M1 for correct equation in $a$ and $b$ for an area (need not be simplified). 2nd M1 for correct equation in $a$ and $b$ using $f(1) = 0 \to a + b = 0$
$f(1) = 0 \to a + b = 0$
Solving simultaneously: $12a - 3a = 1$ OR $15a - 6a = 1$	dM1	3rd dM1 dep on at least 1 M1; for solving two linear equations in $a$ and $b$ by eliminating one variable (allow one slip). If one equation is incorrect we must see explicit method to solve.
$a = \frac{1}{9}$, $b = -\frac{1}{9}$	A1, A1	1st A1 for $a$ and 2nd A1 for $b$ (allow exact equivalents)
(c) 1st M1 for attempt to integrate $ax + b$ with correct limits or $c$ and $F(1) = 0$ or $F(4) = 0.5$	M1, M1	$\int_1^4 \left(\frac{1}{3}t - \frac{1}{9}\right) dt$ or $\int \left(\frac{1}{3}x - \frac{1}{9}\right) dx$ with $c$ and $F(1) = 0$ or $F(4) = 0.5$
2nd M1 for attempt to integrate $\frac{2}{3} - \frac{1}{4}x$ with correct limits & $0.5$ or $+c$ and $F(6) = 1$ or $F(4) = 0.5$	M1	or $\int \left(\frac{2}{3} - \frac{1}{4}x\right) dx$ with $c$ and $F(6) = 1$ or $F(4) = 0.5$
B1 for 1st and 4th line correct	B1	Allow $<$ or $\leq$ anywhere for the last 3 marks
$F(x) = \begin{cases} 0 & x < 1 \\ \frac{1}{18}x^2 - \frac{1}{9}x + \frac{1}{18} & 1 \leq x < 4 \\ \frac{2}{3}x - \frac{1}{8}x^2 - \frac{7}{2} & 4 \leq x \leq 6 \\ 1 & x > 6 \end{cases}$	A1, A1	1st A1 correct 2nd line with limits NB $\frac{1}{18}(x-1)^2$. 2nd A1 correct 3rd line with limits NB $1 - \frac{1}{8}(x-6)^2$

Total: 12

**(a)** Using area of triangle: $P(X > 4) = \frac{(6-4) \times f(4)}{2} = \frac{1}{2}$ [so median of $X$ is 4] OR Using integration: $\int_4^6 \left(\frac{1}{3} - \frac{1}{4}x\right) dx = \left[\frac{1}{3}x - \frac{1}{8}x^2\right]_4 = \frac{1}{2}$ [so median of $X$ is 4] | M1, A1 cso | M1 for a correct expression for area above 4 using triangle or correct integral & attempt to integrate. A1 cso for $\frac{1}{2}$ or $x = 4$ with no errors. NB $f(4) = 1.5 - 1 = 0.5$ or $1.5 - 0.25x = 0.5$ is M0A0

**(b)** Area of triangle from 1 < $x$ < 4 = 0.5: $\frac{(4-1) \times (4a + b)}{2} = 0.5 \to 12a + 3b = 1$ OR $\int_1^4 (ax + b) dx = 0.5 \to 15a + 6b = 1$ | M1, M1 | 1st M1 for correct equation in $a$ and $b$ for an area (need not be simplified). 2nd M1 for correct equation in $a$ and $b$ using $f(1) = 0 \to a + b = 0$

| $f(1) = 0 \to a + b = 0$ | | 

| Solving simultaneously: $12a - 3a = 1$ OR $15a - 6a = 1$ | dM1 | 3rd dM1 dep on at least 1 M1; for solving two linear equations in $a$ and $b$ by eliminating one variable (allow one slip). If one equation is incorrect we must see explicit method to solve.

| $a = \frac{1}{9}$, $b = -\frac{1}{9}$ | A1, A1 | 1st A1 for $a$ and 2nd A1 for $b$ (allow exact equivalents)

**(c)** 1st M1 for attempt to integrate $ax + b$ with correct limits or $c$ and $F(1) = 0$ or $F(4) = 0.5$ | M1, M1 | $\int_1^4 \left(\frac{1}{3}t - \frac{1}{9}\right) dt$ or $\int \left(\frac{1}{3}x - \frac{1}{9}\right) dx$ with $c$ and $F(1) = 0$ or $F(4) = 0.5$

| 2nd M1 for attempt to integrate $\frac{2}{3} - \frac{1}{4}x$ with correct limits & $0.5$ or $+c$ and $F(6) = 1$ or $F(4) = 0.5$ | M1 | or $\int \left(\frac{2}{3} - \frac{1}{4}x\right) dx$ with $c$ and $F(6) = 1$ or $F(4) = 0.5$

| B1 for 1st and 4th line correct | B1 | Allow $<$ or $\leq$ anywhere for the last 3 marks

| $F(x) = \begin{cases} 0 & x < 1 \\ \frac{1}{18}x^2 - \frac{1}{9}x + \frac{1}{18} & 1 \leq x < 4 \\ \frac{2}{3}x - \frac{1}{8}x^2 - \frac{7}{2} & 4 \leq x \leq 6 \\ 1 & x > 6 \end{cases}$ | A1, A1 | 1st A1 correct 2nd line with limits NB $\frac{1}{18}(x-1)^2$. 2nd A1 correct 3rd line with limits NB $1 - \frac{1}{8}(x-6)^2$

**Total: 12**

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Show LaTeX source

3. The random variable $X$ has probability density function given by

$$f ( x ) = \begin{cases} a x + b & 1 \leqslant x < 4 \\ \frac { 3 } { 2 } - \frac { 1 } { 4 } x & 4 \leqslant x \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$

as shown in Figure 1, where $a$ and $b$ are constants.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1a1534ea-4c62-4945-850a-9460ea87fd64-08_634_1132_694_397}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the median of $X$ is 4
\item Find the value of $a$ and the value of $b$
\item Specify fully the cumulative distribution function of $X$
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2017 Q3 [12]}}

This paper (7 questions)

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Q1 11 Q2 12 Q3 12 Q4 13 Q5 11 Q6 7 Q7 9

Answer	Marks	Guidance
(a) Using area of triangle: \(P(X > 4) = \frac{(6-4) \times f(4)}{2} = \frac{1}{2}\) [so median of \(X\) is 4] OR Using integration: \(\int_4^6 \left(\frac{1}{3} - \frac{1}{4}x\right) dx = \left[\frac{1}{3}x - \frac{1}{8}x^2\right]_4 = \frac{1}{2}\) [so median of \(X\) is 4]	M1, A1 cso	M1 for a correct expression for area above 4 using triangle or correct integral & attempt to integrate. A1 cso for \(\frac{1}{2}\) or \(x = 4\) with no errors. NB \(f(4) = 1.5 - 1 = 0.5\) or \(1.5 - 0.25x = 0.5\) is M0A0
(b) Area of triangle from 1 < \(x\) < 4 = 0.5: \(\frac{(4-1) \times (4a + b)}{2} = 0.5 \to 12a + 3b = 1\) OR \(\int_1^4 (ax + b) dx = 0.5 \to 15a + 6b = 1\)	M1, M1	1st M1 for correct equation in \(a\) and \(b\) for an area (need not be simplified). 2nd M1 for correct equation in \(a\) and \(b\) using \(f(1) = 0 \to a + b = 0\)
\(f(1) = 0 \to a + b = 0\)
Solving simultaneously: \(12a - 3a = 1\) OR \(15a - 6a = 1\)	dM1	3rd dM1 dep on at least 1 M1; for solving two linear equations in \(a\) and \(b\) by eliminating one variable (allow one slip). If one equation is incorrect we must see explicit method to solve.
\(a = \frac{1}{9}\), \(b = -\frac{1}{9}\)	A1, A1	1st A1 for \(a\) and 2nd A1 for \(b\) (allow exact equivalents)
(c) 1st M1 for attempt to integrate \(ax + b\) with correct limits or \(c\) and \(F(1) = 0\) or \(F(4) = 0.5\)	M1, M1	\(\int_1^4 \left(\frac{1}{3}t - \frac{1}{9}\right) dt\) or \(\int \left(\frac{1}{3}x - \frac{1}{9}\right) dx\) with \(c\) and \(F(1) = 0\) or \(F(4) = 0.5\)
2nd M1 for attempt to integrate \(\frac{2}{3} - \frac{1}{4}x\) with correct limits & \(0.5\) or \(+c\) and \(F(6) = 1\) or \(F(4) = 0.5\)	M1	or \(\int \left(\frac{2}{3} - \frac{1}{4}x\right) dx\) with \(c\) and \(F(6) = 1\) or \(F(4) = 0.5\)
B1 for 1st and 4th line correct	B1	Allow \(<\) or \(\leq\) anywhere for the last 3 marks
\(F(x) = \begin{cases} 0 & x < 1 \\ \frac{1}{18}x^2 - \frac{1}{9}x + \frac{1}{18} & 1 \leq x < 4 \\ \frac{2}{3}x - \frac{1}{8}x^2 - \frac{7}{2} & 4 \leq x \leq 6 \\ 1 & x > 6 \end{cases}\)	A1, A1	1st A1 correct 2nd line with limits NB \(\frac{1}{18}(x-1)^2\). 2nd A1 correct 3rd line with limits NB \(1 - \frac{1}{8}(x-6)^2\)