Question 3 - A-Level Maths

Edexcel S2 2023 June — Question 3 9 marks

Exam Board	Edexcel
Module	S2 (Statistics 2)
Year	2023
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Calculate probability P(X in interval)
Difficulty	Standard +0.3 This is a standard S2 probability density function question requiring routine integration to find c, calculation of a probability, and understanding of pdf properties. Part (d)-(e) about the mode adds slight conceptual depth (recognizing that f'(x)=0 gives a minimum, not maximum), but overall this is a straightforward multi-part question testing basic pdf concepts with no novel problem-solving required.
Spec	5.03a Continuous random variables: pdf and cdf 5.03f Relate pdf-cdf: medians and percentiles

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

$$f ( x ) = \left\{ \begin{array} { c c } \frac { 1 } { 48 } \left( x ^ { 2 } - 8 x + c \right) & 2 \leqslant x \leqslant 5 \\ 0 & \text { otherwise } \end{array} \right.$$

Show that $c = 31$
Find $\mathrm { P } ( 2 < X < 3 )$
State whether the lower quartile of $X$ is less than 3, equal to 3 or greater than 3 Give a reason for your answer. Kei does the following to work out the mode of $X$ $$\begin{aligned} f ^ { \prime } ( x ) & = \frac { 1 } { 48 } ( 2 x - 8 ) \\ 0 & = \frac { 1 } { 48 } ( 2 x - 8 ) \\ x & = 4 \end{aligned}$$ Hence the mode of $X$ is 4 Kei's answer for the mode is incorrect.
Explain why Kei's method does not give the correct value for the mode.
Find the mode of $X$ Give a reason for your answer.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) $\int_2^5 \frac{1}{48}\left(x^2 - 8x + c\right)dx = 1$	M1	1st M1 Setting up integral and equating to 1 (condone missing dx) limits not needed
$1 = \frac{1}{48}\left[\frac{x^3}{3} - 4x^2 + cx\right]_2$	M1	2nd M1 attempting to integrate $f(x)$ at least one term $x^n \to x^{n+1}$ (need not be = 1)
$1 = \frac{1}{48}\left(\left(\frac{5^3}{3} - 4(5^2) + 5c\right) - \left(\frac{2^3}{3} - 4(2^2) + 2c\right)\right)$ or $48 = 39 - 84 + 3c$	A1 cso	Use of integration of $f(x)$ with $F(2) = 0$ and $F(5) = 1$ can score M1M1. A1* cso including use of correct limits. There should be at least one line of working between scoring the 2nd M1 and arriving at the given answer. Allow a verification method 1st M1 setting up integral 2nd M1 A1 cso use of correct limits to show that it integrates to 1 and concluding that $c = 31$
$(\Rightarrow 3c = 93 \Rightarrow) c = 31$		(3)
(b) $P(2 < X < 3) = \frac{1}{48}\left[\frac{x^3}{3} - 4x^2 + 31x\right]_2$	M1	M1 for use of integration of $f(x)$ $x^n \to x^{n+1}$ with correct limits 2 and 3 (ff from their (a))
$\frac{1}{48}\left(\left(\frac{3^3}{3} - 4(3^2) + 31(3)\right) - \left(\frac{2^3}{3} - 4(2^2) + 31(2)\right)\right) = \frac{13}{36}$ (=awrt 0.361)	A1	A1 allow awrt 0.361 (correct answer scores 2 out of 2)
(2)
(c) Less than 3 since $\frac{13}{36} > 0.25$	B1	B1 less than 3 with correct reasoning. May use their part (b), but must be consistent with 'less than 3'. If the lower quartile is found awrt 2.67, allow LQ/2.67 < 3
(d) $x = 4$ leads to the minimum/lowest value of $f(x) / f(x)$ is a positive quadratic	B1	B1 correct reason why the method does not give the correct mode. Allow a sketch of $f(x)$. Also allow, e.g. 'Kei's method did not consider the end-points'
(e) Considers $x = 2$ and $x = 5$ by e.g. $f(2) = 0.39(583)\left[= \frac{39}{48}\right]$ and $f(5) = 0.3\left[= \frac{16}{48}\right]$ (so $f(2) > f(5)$)	M1	M1 considers end-points. Sketch of $f(x)$ from $x = 2$ to $x = 5$. $x = 2$ is further than $x = 4$ (then $x = 5$). Mode is $x = 2$
Mode is $x = 2$	A1	A1 mode is 2 cao Answer only scores M0A0. Must have some justification.
(2)
[9 marks]

**(a)** $\int_2^5 \frac{1}{48}\left(x^2 - 8x + c\right)dx = 1$ | M1 | 1st M1 Setting up integral and equating to 1 (condone missing dx) limits not needed

$1 = \frac{1}{48}\left[\frac{x^3}{3} - 4x^2 + cx\right]_2$ | M1 | 2nd M1 attempting to integrate $f(x)$ at least one term $x^n \to x^{n+1}$ (need not be = 1)

$1 = \frac{1}{48}\left(\left(\frac{5^3}{3} - 4(5^2) + 5c\right) - \left(\frac{2^3}{3} - 4(2^2) + 2c\right)\right)$ or $48 = 39 - 84 + 3c$ | A1 cso | Use of integration of $f(x)$ with $F(2) = 0$ and $F(5) = 1$ can score M1M1. A1* cso including use of correct limits. There should be at least one line of working between scoring the 2nd M1 and arriving at the given answer. Allow a verification method 1st M1 setting up integral 2nd M1 A1 cso use of correct limits to show that it integrates to 1 and concluding that $c = 31$

$(\Rightarrow 3c = 93 \Rightarrow) c = 31$ | | (3)

**(b)** $P(2 < X < 3) = \frac{1}{48}\left[\frac{x^3}{3} - 4x^2 + 31x\right]_2$ | M1 | M1 for use of integration of $f(x)$ $x^n \to x^{n+1}$ with correct limits 2 and 3 (ff from their (a))

$\frac{1}{48}\left(\left(\frac{3^3}{3} - 4(3^2) + 31(3)\right) - \left(\frac{2^3}{3} - 4(2^2) + 31(2)\right)\right) = \frac{13}{36}$ (=awrt 0.361) | A1 | A1 allow awrt 0.361 (correct answer scores 2 out of 2)

| | (2)

**(c)** Less than 3 since $\frac{13}{36} > 0.25$ | B1 | B1 less than 3 with correct reasoning. May use their part (b), but must be consistent with 'less than 3'. If the lower quartile is found awrt 2.67, allow LQ/2.67 < 3

**(d)** $x = 4$ leads to the minimum/lowest value of $f(x) / f(x)$ is a positive quadratic | B1 | B1 correct reason why the method does not give the correct mode. Allow a sketch of $f(x)$. Also allow, e.g. 'Kei's method did not consider the end-points'

**(e)** Considers $x = 2$ and $x = 5$ by e.g. $f(2) = 0.39(583)\left[= \frac{39}{48}\right]$ and $f(5) = 0.3\left[= \frac{16}{48}\right]$ (so $f(2) > f(5)$) | M1 | M1 considers end-points. Sketch of $f(x)$ from $x = 2$ to $x = 5$. $x = 2$ is further than $x = 4$ (then $x = 5$). Mode is $x = 2$

Mode is $x = 2$ | A1 | A1 mode is 2 cao Answer only scores M0A0. Must have some justification.

| | (2)

| | [9 marks]

Show LaTeX source

\begin{enumerate}
  \item The continuous random variable $X$ has probability density function given by
\end{enumerate}

$$f ( x ) = \left\{ \begin{array} { c c } 
\frac { 1 } { 48 } \left( x ^ { 2 } - 8 x + c \right) & 2 \leqslant x \leqslant 5 \\
0 & \text { otherwise }
\end{array} \right.$$

(a) Show that $c = 31$\\
(b) Find $\mathrm { P } ( 2 < X < 3 )$\\
(c) State whether the lower quartile of $X$ is less than 3, equal to 3 or greater than 3 Give a reason for your answer.

Kei does the following to work out the mode of $X$

$$\begin{aligned}
f ^ { \prime } ( x ) & = \frac { 1 } { 48 } ( 2 x - 8 ) \\
0 & = \frac { 1 } { 48 } ( 2 x - 8 ) \\
x & = 4
\end{aligned}$$

Hence the mode of $X$ is 4

Kei's answer for the mode is incorrect.\\
(d) Explain why Kei's method does not give the correct value for the mode.\\
(e) Find the mode of $X$

Give a reason for your answer.

\hfill \mbox{\textit{Edexcel S2 2023 Q3 [9]}}

This paper (7 questions)

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Q1 11 Q2 4 Q3 9 Q4 13 Q5 14 Q6 12 Q7 12

Answer	Marks	Guidance
(a) \(\int_2^5 \frac{1}{48}\left(x^2 - 8x + c\right)dx = 1\)	M1	1st M1 Setting up integral and equating to 1 (condone missing dx) limits not needed
\(1 = \frac{1}{48}\left[\frac{x^3}{3} - 4x^2 + cx\right]_2\)	M1	2nd M1 attempting to integrate \(f(x)\) at least one term \(x^n \to x^{n+1}\) (need not be = 1)
\(1 = \frac{1}{48}\left(\left(\frac{5^3}{3} - 4(5^2) + 5c\right) - \left(\frac{2^3}{3} - 4(2^2) + 2c\right)\right)\) or \(48 = 39 - 84 + 3c\)	A1 cso	Use of integration of \(f(x)\) with \(F(2) = 0\) and \(F(5) = 1\) can score M1M1. A1* cso including use of correct limits. There should be at least one line of working between scoring the 2nd M1 and arriving at the given answer. Allow a verification method 1st M1 setting up integral 2nd M1 A1 cso use of correct limits to show that it integrates to 1 and concluding that \(c = 31\)
\((\Rightarrow 3c = 93 \Rightarrow) c = 31\)		(3)
(b) \(P(2 < X < 3) = \frac{1}{48}\left[\frac{x^3}{3} - 4x^2 + 31x\right]_2\)	M1	M1 for use of integration of \(f(x)\) \(x^n \to x^{n+1}\) with correct limits 2 and 3 (ff from their (a))
\(\frac{1}{48}\left(\left(\frac{3^3}{3} - 4(3^2) + 31(3)\right) - \left(\frac{2^3}{3} - 4(2^2) + 31(2)\right)\right) = \frac{13}{36}\) (=awrt 0.361)	A1	A1 allow awrt 0.361 (correct answer scores 2 out of 2)
(2)
(c) Less than 3 since \(\frac{13}{36} > 0.25\)	B1	B1 less than 3 with correct reasoning. May use their part (b), but must be consistent with 'less than 3'. If the lower quartile is found awrt 2.67, allow LQ/2.67 < 3
(d) \(x = 4\) leads to the minimum/lowest value of \(f(x) / f(x)\) is a positive quadratic	B1	B1 correct reason why the method does not give the correct mode. Allow a sketch of \(f(x)\). Also allow, e.g. 'Kei's method did not consider the end-points'
(e) Considers \(x = 2\) and \(x = 5\) by e.g. \(f(2) = 0.39(583)\left[= \frac{39}{48}\right]\) and \(f(5) = 0.3\left[= \frac{16}{48}\right]\) (so \(f(2) > f(5)\))	M1	M1 considers end-points. Sketch of \(f(x)\) from \(x = 2\) to \(x = 5\). \(x = 2\) is further than \(x = 4\) (then \(x = 5\)). Mode is \(x = 2\)
Mode is \(x = 2\)	A1	A1 mode is 2 cao Answer only scores M0A0. Must have some justification.
(2)
[9 marks]