Question 6 - A-Level Maths

Edexcel FS2 2023 June — Question 6 10 marks

Exam Board	Edexcel
Module	FS2 (Further Statistics 2)
Year	2023
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Cumulative distribution functions
Type	CDF with additional constraints
Difficulty	Challenging +1.2 This is a Further Maths Statistics question requiring systematic application of CDF properties (continuity at x=4, using P(X<2)=2/3) to find constants, then standard techniques (solving F(x)=0.5 for median, differentiating for pdf, finding stationary points for mode). While it involves multiple parts and algebraic manipulation, each step follows directly from definitions without requiring novel insight. The Further Maths context and multi-step nature place it above average difficulty.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf

The continuous random variable $X$ has cumulative distribution function given by

$$F ( x ) = \left\{ \begin{array} { c r } 0 & x < 0 \\ k \left( x - a x ^ { 2 } \right) & 0 \leqslant x \leqslant 4 \\ 1 & x > 4 \end{array} \right.$$ The values of $a$ and $k$ are positive constants such that $\mathrm { P } ( X < 2 ) = \frac { 2 } { 3 }$

Find the exact value of the median of $X$
Find the probability density function of $X$
Hence, deduce the value of the mode of $X$, giving a reason for your answer.

Show mark scheme Show mark scheme source

Question 6:

Part (a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$F(2) = \frac{2}{3}$ and $F(4) = 1$ used: $k(2-4a) = \frac{2}{3}$, $k(4-16a) = 1$	M1	2.1 - using both conditions to form two expressions in $a$ and $k$
$\frac{2}{3(2-4a)} = \frac{1}{(4-16a)} \rightarrow a = \frac{1}{10}$, $k = \frac{5}{12}$	M1A1	1.1b - solving simultaneously; both values correct
$\frac{5}{12}\left(m - \frac{1}{10}m^2\right) = 0.5 \rightarrow m^2 - 10m + 12 = 0$	M1M1	3.1a, 1.1b - setting up $F(m)=0.5$; forming 3TQ
$m = 5 - \sqrt{13}$ (reject $m = 5 + \sqrt{13}$)	A1	1.1b - selecting correct value

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$f(x) = \frac{d}{dx}(F(x))$	M1	1.1b - differentiating $F(x)$, at least one term correct for their $a$ and $k$
$f(x) = \begin{cases} \frac{5}{12}(1 - 0.2x) & 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$	A1ft	1.1b - correct ft expression with correct limits

Part (c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Mode is $X = 0$	B1	2.2a
...since $f(x)$ is linear with negative gradient, or $f(x)$ is a decreasing function	dB1	2.4 - correct reasoning; may be shown via diagram of linear function with negative gradient, intersecting $y$-axis, never negative in $y$

# Question 6:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(2) = \frac{2}{3}$ and $F(4) = 1$ used: $k(2-4a) = \frac{2}{3}$, $k(4-16a) = 1$ | M1 | 2.1 - using both conditions to form two expressions in $a$ and $k$ |
| $\frac{2}{3(2-4a)} = \frac{1}{(4-16a)} \rightarrow a = \frac{1}{10}$, $k = \frac{5}{12}$ | M1A1 | 1.1b - solving simultaneously; both values correct |
| $\frac{5}{12}\left(m - \frac{1}{10}m^2\right) = 0.5 \rightarrow m^2 - 10m + 12 = 0$ | M1M1 | 3.1a, 1.1b - setting up $F(m)=0.5$; forming 3TQ |
| $m = 5 - \sqrt{13}$ (reject $m = 5 + \sqrt{13}$) | A1 | 1.1b - selecting correct value |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = \frac{d}{dx}(F(x))$ | M1 | 1.1b - differentiating $F(x)$, at least one term correct for their $a$ and $k$ |
| $f(x) = \begin{cases} \frac{5}{12}(1 - 0.2x) & 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$ | A1ft | 1.1b - correct ft expression with correct limits |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Mode is $X = 0$ | B1 | 2.2a |
| ...since $f(x)$ is linear with negative gradient, or $f(x)$ is a decreasing function | dB1 | 2.4 - correct reasoning; may be shown via diagram of linear function with negative gradient, intersecting $y$-axis, never negative in $y$ |

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Show LaTeX source

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

$$F ( x ) = \left\{ \begin{array} { c r } 
0 & x < 0 \\
k \left( x - a x ^ { 2 } \right) & 0 \leqslant x \leqslant 4 \\
1 & x > 4
\end{array} \right.$$

The values of $a$ and $k$ are positive constants such that $\mathrm { P } ( X < 2 ) = \frac { 2 } { 3 }$\\
(a) Find the exact value of the median of $X$\\
(b) Find the probability density function of $X$\\
(c) Hence, deduce the value of the mode of $X$, giving a reason for your answer.

\hfill \mbox{\textit{Edexcel FS2 2023 Q6 [10]}}

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