Edexcel FS2 AS 2024 June — Question 1 8 marks

Exam BoardEdexcel
ModuleFS2 AS (Further Statistics 2 AS)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypePDF from CDF
DifficultyStandard +0.3 This is a straightforward FS2 question requiring differentiation of a piecewise CDF to find the PDF, sketching it, identifying skewness from symmetry, and solving an equation using the CDF. All steps are routine applications of standard techniques with no novel insight required, making it slightly easier than average for A-level.
Spec5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration
  1. A continuous random variable \(X\) has cumulative distribution function \(\mathrm { F } ( x )\) given by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c r } 0 & x < - 1 \\ \frac { 1 } { 5 } ( x + 1 ) ^ { 2 } & - 1 \leqslant x \leqslant 0 \\ 1 - \frac { 1 } { 20 } ( 4 - x ) ^ { 2 } & 0 < x \leqslant 4 \\ 1 & x > 4 \end{array} \right.$$
  1. Find the probability density function, \(\mathrm { f } ( x )\)
    1. Sketch \(\mathrm { f } ( x )\)
    2. Hence describe the skewness of the distribution.
  3. Find, to 3 significant figures, the value of \(c\) such that $$\mathrm { P } ( 1 < X < c ) = \mathrm { P } ( c < X < 2 )$$
Question 1:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d}{dx}\left[\frac{1}{5}(x+1)^2\right] = \frac{2}{5}(x+1)\) or \(\frac{d}{dx}\left[1-\frac{1}{20}(4-x)^2\right] = \frac{1}{10}(4-x)\)M1 Attempt to differentiate 2nd line of cdf in form \(k(x+1)\) or 3rd line in form \(m(4-x)\)
\(f(x) = \begin{cases} \frac{2}{5}(x+1) & -1 \leq x \leq 0 \\ \frac{1}{10}(4-x) & 0 < x \leq 4 \\ 0 & \text{otherwise} \end{cases}\)A1 Correct pdf. Condone missing "0 otherwise" and mis-use of \(<\) or \(\leq\)
Part (b)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Shape: Triangle with longer slant in 1st quadrantB1 Correct shape; triangle must have both ends on axis
Labels: \(-1\), \(4\) on horizontal axis and \(\frac{2}{5}\) on vertical axisB1 Correct labels; all three needed
Part (b)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Probability density function has a longer tail to right, so there is positive skewdB1 Correct description of positive skew. Dependent on sketch that clearly shows positive skew
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(1 < X < 2) = F(2) - F(1)\) or \(2F(c) = F(2) + F(1)\)
\(= \left(1 - \frac{4}{20}\right) - \left(1 - \frac{9}{20}\right) = \frac{5}{20}\) or \(\frac{1}{4}\) or stating \(F(c) = 0.675\)M1 Suitable start e.g. showing \(P(1 < X < 2) = 0.25\); e.g. \(F(c) = 0.675\) or \(2F(c) = 0.8 + 0.55\)
\(P(c < X < 2) = \frac{1}{8} \Rightarrow \frac{1}{8} = \frac{(4-c)^2}{20} - \frac{4}{20}\)M1 Rearranging to form quadratic equation in \(c\); e.g. \(2c^2 - 16c + 19 = 0\). Allow even if wrong part of \(F(x)\) used so M0M1A0 is possible
\(c = \frac{8 - \sqrt{26}}{2}\) or \(1.45049\ldots\) awrt \(\mathbf{1.45}\)A1 For awrt 1.45 only. NB other root is \(6.549\ldots\) and if not rejected score A0 
# Question 1:

## Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}\left[\frac{1}{5}(x+1)^2\right] = \frac{2}{5}(x+1)$ or $\frac{d}{dx}\left[1-\frac{1}{20}(4-x)^2\right] = \frac{1}{10}(4-x)$ | M1 | Attempt to differentiate 2nd line of cdf in form $k(x+1)$ or 3rd line in form $m(4-x)$ |
| $f(x) = \begin{cases} \frac{2}{5}(x+1) & -1 \leq x \leq 0 \\ \frac{1}{10}(4-x) & 0 < x \leq 4 \\ 0 & \text{otherwise} \end{cases}$ | A1 | Correct pdf. Condone missing "0 otherwise" and mis-use of $<$ or $\leq$ |

## Part (b)(i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Shape: Triangle with longer slant in 1st quadrant | B1 | Correct shape; triangle must have both ends on axis |
| Labels: $-1$, $4$ on horizontal axis and $\frac{2}{5}$ on vertical axis | B1 | Correct labels; all three needed |

## Part (b)(ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Probability density function has a longer tail to right, so there is **positive skew** | dB1 | Correct description of positive skew. Dependent on sketch that clearly shows positive skew |

## Part (c)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(1 < X < 2) = F(2) - F(1)$ or $2F(c) = F(2) + F(1)$ | | |
| $= \left(1 - \frac{4}{20}\right) - \left(1 - \frac{9}{20}\right) = \frac{5}{20}$ or $\frac{1}{4}$ or stating $F(c) = 0.675$ | M1 | Suitable start e.g. showing $P(1 < X < 2) = 0.25$; e.g. $F(c) = 0.675$ or $2F(c) = 0.8 + 0.55$ |
| $P(c < X < 2) = \frac{1}{8} \Rightarrow \frac{1}{8} = \frac{(4-c)^2}{20} - \frac{4}{20}$ | M1 | Rearranging to form quadratic equation in $c$; e.g. $2c^2 - 16c + 19 = 0$. Allow even if wrong part of $F(x)$ used so M0M1A0 is possible |
| $c = \frac{8 - \sqrt{26}}{2}$ or $1.45049\ldots$ awrt $\mathbf{1.45}$ | A1 | For awrt 1.45 only. NB other root is $6.549\ldots$ and if not rejected score A0 |

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\begin{enumerate}
  \item A continuous random variable $X$ has cumulative distribution function $\mathrm { F } ( x )$ given by
\end{enumerate}

$$\mathrm { F } ( x ) = \left\{ \begin{array} { c r } 
0 & x < - 1 \\
\frac { 1 } { 5 } ( x + 1 ) ^ { 2 } & - 1 \leqslant x \leqslant 0 \\
1 - \frac { 1 } { 20 } ( 4 - x ) ^ { 2 } & 0 < x \leqslant 4 \\
1 & x > 4
\end{array} \right.$$

(a) Find the probability density function, $\mathrm { f } ( x )$\\
(b) (i) Sketch $\mathrm { f } ( x )$\\
(ii) Hence describe the skewness of the distribution.\\
(c) Find, to 3 significant figures, the value of $c$ such that

$$\mathrm { P } ( 1 < X < c ) = \mathrm { P } ( c < X < 2 )$$

\hfill \mbox{\textit{Edexcel FS2 AS 2024 Q1 [8]}}
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