Question 5 - A-Level Maths

Edexcel S2 2024 October — Question 5

Exam Board	Edexcel
Module	S2 (Statistics 2)
Year	2024
Session	October
Paper	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Find or specify CDF
Difficulty	Moderate -0.3 This is a standard S2 question on piecewise probability density functions requiring integration to find the CDF and probability calculations. Part (a) is guided (show that), parts (b) and (c) involve routine integration and substitution with no conceptual challenges beyond careful bookkeeping of the piecewise cases.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03e Find cdf: by integration

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

$$f ( x ) = \begin{cases} \frac { 1 } { 4 } ( 3 - x ) & 1 \leqslant x \leqslant 2 \\ \frac { 1 } { 4 } & 2 < x \leqslant 3 \\ \frac { 1 } { 4 } ( x - 2 ) & 3 < x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ The cumulative distribution function of $X$ is $\mathrm { F } ( x )$

Show that $\mathrm { F } ( x ) = \frac { 1 } { 4 } \left( 3 x - \frac { x ^ { 2 } } { 2 } \right) - \frac { 5 } { 8 }$ for $1 \leqslant x \leqslant 2$
Find $\mathrm { F } ( x )$ for all values of $x$
Find $\mathrm { P } ( 1.2 < X < 3.1 )$

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a)	$\int_1^4 \frac{1}{4}(3-t) dt = \frac{1}{4}\left[3t - \frac{t^2}{2}\right]_1$ or $\int_1^4(3-x) dx = \frac{1}{4}\left[3x - \frac{x^2}{2}\right] + C$	M1
$\frac{1}{4}\left[\left(3x - \frac{x^2}{2}\right) - \left(3 - \frac{1}{2}\right)\right]$ or $\frac{1}{4}\left[3(1) - \frac{(1)^2}{2}\right] + C = 0$ and $C = -\frac{5}{8}$	A1*
Leading to $\frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}$ [for $1_r, x_r, 2$] *		(2)
(b)	$\int_2 \frac{1}{4} dt + F(2)$ or $\int_1 \frac{1}{4} dx$ and using $+c$ with $F(2) = \frac{3}{8}$ or $0.25(x-2) + F(2)$	M1
$\int_1 \frac{1}{4}(t-2) dt + F(3)$ or $\int_1 \frac{1}{4}(x-2) dx$ and using $+c$ with either $F(3) = \frac{5}{8}$ or $F(4) = 1$	M1
A1
A1
B1	(5)
(c)	$P(1.2 < \bar{X} < 3.1) = F(3.1) - F(1.2)$	M1 A1
$\left(\frac{1}{4}\left(\frac{(3.1)^2}{2} - 2(3.1)\right) + 1\right) - \left(\frac{1}{4}\left(3(1.2) - \frac{(1.2)^2}{2}\right) - \frac{5}{8}\right) = \frac{89}{160}$ awrt 0.556		(2)
Total	9

Notes for Question 5:

- (a) M1: For a correct method for $1_r, x_r, 2$ Condone poor notation e.g. $\int_1 \frac{1}{4}(3-x) dx$

- **(a) A1*:** A fully correct solution with substitution seen or $C$ found leading to $F(x) = \frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}$

- (b) M1: For a correct method for $2 < x_r, 3$

- (b) M1: For a correct method for $3 < x_r, 4$

- (b) A1: Third line correct including inequality. Allow $<$ instead of $\leq$

- (b) A1: Fourth line correct including inequality. Allow $<$ instead of $\leq$

- (b) B1: First and fifth line correct. Allow "otherwise" for the range on the first or fifth line but not both

- For use of $F(3.1) - F(1.2)$ from the correct lines of their $F(x)$ allow ft on their 4th line

- (c) M1: For use of $f(x)$ or area e.g. $\frac{1}{2} \times \frac{7}{10} \times 0.8 + 1 \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{4} \times 0.1$

- (c) A1: For $\frac{89}{160}$ oe or awrt 0.556 NB: Use of $F(3.1)$ with $\frac{1}{8}(2x-1)$ for $2 < x_r, 3$ gives 0.555 and scores M0A0

| (a) | $\int_1^4 \frac{1}{4}(3-t) dt = \frac{1}{4}\left[3t - \frac{t^2}{2}\right]_1$ or $\int_1^4(3-x) dx = \frac{1}{4}\left[3x - \frac{x^2}{2}\right] + C$ | M1 |  |
| | $\frac{1}{4}\left[\left(3x - \frac{x^2}{2}\right) - \left(3 - \frac{1}{2}\right)\right]$ or $\frac{1}{4}\left[3(1) - \frac{(1)^2}{2}\right] + C = 0$ and $C = -\frac{5}{8}$ | A1* |  |
| | Leading to $\frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}$ [for $1_r, x_r, 2$] * | | (2) |
| (b) | $\int_2 \frac{1}{4} dt + F(2)$ or $\int_1 \frac{1}{4} dx$ and using $+c$ with $F(2) = \frac{3}{8}$ or $0.25(x-2) + F(2)$ | M1 |  |
| | $\int_1 \frac{1}{4}(t-2) dt + F(3)$ or $\int_1 \frac{1}{4}(x-2) dx$ and using $+c$ with either $F(3) = \frac{5}{8}$ or $F(4) = 1$ | M1 |  |
| | | A1 |  |
| | | A1 |  |
| | | B1 | (5) |
| (c) | $P(1.2 < \bar{X} < 3.1) = F(3.1) - F(1.2)$ | M1 A1 |  |
| | $\left(\frac{1}{4}\left(\frac{(3.1)^2}{2} - 2(3.1)\right) + 1\right) - \left(\frac{1}{4}\left(3(1.2) - \frac{(1.2)^2}{2}\right) - \frac{5}{8}\right) = \frac{89}{160}$ awrt 0.556 | | (2) |
| | **Total** | **9** |  |

**Notes for Question 5:**

- **(a) M1:** For a correct method for $1_r, x_r, 2$ Condone poor notation e.g. $\int_1 \frac{1}{4}(3-x) dx$
- **(a) A1*:** A fully correct solution with substitution seen or $C$ found leading to $F(x) = \frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}$
- **(b) M1:** For a correct method for $2 < x_r, 3$
- **(b) M1:** For a correct method for $3 < x_r, 4$
- **(b) A1:** Third line correct including inequality. Allow $<$ instead of $\leq$
- **(b) A1:** Fourth line correct including inequality. Allow $<$ instead of $\leq$
- **(b) B1:** First and fifth line correct. Allow "otherwise" for the range on the first or fifth line but not both
  - For use of $F(3.1) - F(1.2)$ from the correct lines of their $F(x)$ allow ft on their 4th line
- **(c) M1:** For use of $f(x)$ or area e.g. $\frac{1}{2} \times \frac{7}{10} \times 0.8 + 1 \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{4} \times 0.1$
- **(c) A1:** For $\frac{89}{160}$ oe or awrt 0.556 NB: Use of $F(3.1)$ with $\frac{1}{8}(2x-1)$ for $2 < x_r, 3$ gives 0.555 and scores M0A0

---

Show LaTeX source

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

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

The cumulative distribution function of $X$ is $\mathrm { F } ( x )$\\
(a) Show that $\mathrm { F } ( x ) = \frac { 1 } { 4 } \left( 3 x - \frac { x ^ { 2 } } { 2 } \right) - \frac { 5 } { 8 }$ for $1 \leqslant x \leqslant 2$\\
(b) Find $\mathrm { F } ( x )$ for all values of $x$\\
(c) Find $\mathrm { P } ( 1.2 < X < 3.1 )$

\hfill \mbox{\textit{Edexcel S2 2024 Q5}}

This paper (7 questions)

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Q1 Q2 Q3 Q4 Q5 Q6 Q7

Edexcel S2 2024 October — Question 5

Notes for Question 5:

- (a) M1: For a correct method for \(1_r, x_r, 2\) Condone poor notation e.g. \(\int_1 \frac{1}{4}(3-x) dx\)

- **(a) A1*:** A fully correct solution with substitution seen or \(C\) found leading to \(F(x) = \frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}\)

- (b) M1: For a correct method for \(2 < x_r, 3\)

- (b) M1: For a correct method for \(3 < x_r, 4\)

- (b) A1: Third line correct including inequality. Allow \(<\) instead of \(\leq\)

- (b) A1: Fourth line correct including inequality. Allow \(<\) instead of \(\leq\)

- (b) B1: First and fifth line correct. Allow "otherwise" for the range on the first or fifth line but not both

- For use of \(F(3.1) - F(1.2)\) from the correct lines of their \(F(x)\) allow ft on their 4th line

- (c) M1: For use of \(f(x)\) or area e.g. \(\frac{1}{2} \times \frac{7}{10} \times 0.8 + 1 \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{4} \times 0.1\)

- (c) A1: For \(\frac{89}{160}\) oe or awrt 0.556 NB: Use of \(F(3.1)\) with \(\frac{1}{8}(2x-1)\) for \(2 < x_r, 3\) gives 0.555 and scores M0A0

This paper (7 questions)

Answer	Marks	Guidance
(a)	\(\int_1^4 \frac{1}{4}(3-t) dt = \frac{1}{4}\left[3t - \frac{t^2}{2}\right]_1\) or \(\int_1^4(3-x) dx = \frac{1}{4}\left[3x - \frac{x^2}{2}\right] + C\)	M1
\(\frac{1}{4}\left[\left(3x - \frac{x^2}{2}\right) - \left(3 - \frac{1}{2}\right)\right]\) or \(\frac{1}{4}\left[3(1) - \frac{(1)^2}{2}\right] + C = 0\) and \(C = -\frac{5}{8}\)	A1*
Leading to \(\frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}\) [for \(1_r, x_r, 2\)] *		(2)
(b)	\(\int_2 \frac{1}{4} dt + F(2)\) or \(\int_1 \frac{1}{4} dx\) and using \(+c\) with \(F(2) = \frac{3}{8}\) or \(0.25(x-2) + F(2)\)	M1
\(\int_1 \frac{1}{4}(t-2) dt + F(3)\) or \(\int_1 \frac{1}{4}(x-2) dx\) and using \(+c\) with either \(F(3) = \frac{5}{8}\) or \(F(4) = 1\)	M1
A1
A1
B1	(5)
(c)	\(P(1.2 < \bar{X} < 3.1) = F(3.1) - F(1.2)\)	M1 A1
\(\left(\frac{1}{4}\left(\frac{(3.1)^2}{2} - 2(3.1)\right) + 1\right) - \left(\frac{1}{4}\left(3(1.2) - \frac{(1.2)^2}{2}\right) - \frac{5}{8}\right) = \frac{89}{160}\) awrt 0.556		(2)
Total	9

Edexcel S2 2024 October — Question 5

Notes for Question 5:

- (a) M1: For a correct method for \(1_r, x_r, 2\) Condone poor notation e.g. \(\int_1 \frac{1}{4}(3-x) dx\)

- (a) A1*: A fully correct solution with substitution seen or \(C\) found leading to \(F(x) = \frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}\)

- (b) M1: For a correct method for \(2 < x_r, 3\)

- (b) M1: For a correct method for \(3 < x_r, 4\)

- (b) A1: Third line correct including inequality. Allow \(<\) instead of \(\leq\)

- (b) A1: Fourth line correct including inequality. Allow \(<\) instead of \(\leq\)

- (b) B1: First and fifth line correct. Allow "otherwise" for the range on the first or fifth line but not both

- For use of \(F(3.1) - F(1.2)\) from the correct lines of their \(F(x)\) allow ft on their 4th line

- (c) M1: For use of \(f(x)\) or area e.g. \(\frac{1}{2} \times \frac{7}{10} \times 0.8 + 1 \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{4} \times 0.1\)

- (c) A1: For \(\frac{89}{160}\) oe or awrt 0.556 NB: Use of \(F(3.1)\) with \(\frac{1}{8}(2x-1)\) for \(2 < x_r, 3\) gives 0.555 and scores M0A0

This paper (7 questions)

- **(a) A1*:** A fully correct solution with substitution seen or \(C\) found leading to \(F(x) = \frac{1}{4}\left(3x - \frac{x^2}{2}\right) - \frac{5}{8}\)