Edexcel S2 2009 January — Question 7 13 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2009
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypePDF to CDF derivation
DifficultyStandard +0.3 This is a straightforward S2 question requiring standard integration of a linear pdf to find the cdf, then solving quadratic equations to find quartiles. All techniques are routine: integrate f(x), apply boundary conditions, solve F(x)=0.75 and F(x)=0.25, and compare median/quartiles for skewness. The 'show that' parts guide students through the solution. Slightly easier than average due to the linear pdf making integration trivial and the structured scaffolding.
Spec5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles
A random variable \(X\) has probability density function given by $$\text{f}(x) = \begin{cases} -\frac{2}{9}x + \frac{8}{9} & 1 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}$$
  1. Show that the cumulative distribution function F(x) can be written in the form \(ax^2 + bx + c\), for \(1 \leqslant x \leqslant 4\) where \(a\), \(b\) and \(c\) are constants. [3]
  2. Define fully the cumulative distribution function F(x). [2]
  3. Show that the upper quartile of \(X\) is 2.5 and find the lower quartile. [6]
Given that the median of \(X\) is 1.88
  1. describe the skewness of the distribution. Give a reason for your answer. [2]
Part (a)
AnswerMarks Guidance
\(F(x_0) = \int_1^{x_0} -\frac{2}{9}x + \frac{8}{9}dx = \left[-\frac{1}{9}x^2 + \frac{8}{9}x\right]_1^{x_0}\)M1A1
\(= \left[-\frac{1}{9}x^2 + \frac{8}{9}x\right] - \left[-\frac{1}{9} + \frac{8}{9}\right] = -\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9}\)A1 (3 marks)
Part (b)
AnswerMarks Guidance
\[F(x) = \begin{cases} 0 & x < 1 \\ -\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9} & 1 \leq x \leq 4 \\ 1 & x > 4 \end{cases}\]B1B1/ (2 marks)
Part (c)
AnswerMarks
\(F(x) = 0.75\); or \(F(2.5) = -\frac{1}{9} \times 2.5^2 + \frac{8}{9} \times 2.5 - \frac{7}{9}\)M1;
\(-\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9} = 0.75\)
AnswerMarks Guidance
\(4x^2 - 32x + 55 = 0\) or \(-x^2 + 8x - 13.75 = 0\), \(x = 2.5\)\(= 0.75\) cso
and \(F(x) = 0.25\)
\(-\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9} = 0.25\)
AnswerMarks Guidance
\(-x^2 + 8x - 7 = 2.25\) or \(-x^2 + 8x - 9.25 = 0\)quadratic 3 terms = 0 M1 dep; M1 dep
\(x = \frac{-8 \pm \sqrt{8^2 - 4 \times (-1) \times (-9.25)}}{2 \times (-1)} = 1.40\)A1 (6 marks)
Part (d)
AnswerMarks Guidance
\(Q_3 - Q_2 > Q_2 - Q_1\) Or mode = 1 and mode < median Or mean = 2 and median < mode; Sketch of pdf here or be referred to if in a different part of the question; Box plot with \(Q_1, Q_2, Q_3\) values marked on; Positive skewM1, A1 (2 marks) 
## Part (a)
$F(x_0) = \int_1^{x_0} -\frac{2}{9}x + \frac{8}{9}dx = \left[-\frac{1}{9}x^2 + \frac{8}{9}x\right]_1^{x_0}$ | M1A1 |
$= \left[-\frac{1}{9}x^2 + \frac{8}{9}x\right] - \left[-\frac{1}{9} + \frac{8}{9}\right] = -\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9}$ | A1 | (3 marks)

## Part (b)
$$F(x) = \begin{cases} 0 & x < 1 \\ -\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9} & 1 \leq x \leq 4 \\ 1 & x > 4 \end{cases}$$ | B1B1/ | (2 marks)

## Part (c)
$F(x) = 0.75$; or $F(2.5) = -\frac{1}{9} \times 2.5^2 + \frac{8}{9} \times 2.5 - \frac{7}{9}$ | M1; |
$-\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9} = 0.75$
$4x^2 - 32x + 55 = 0$ or $-x^2 + 8x - 13.75 = 0$, $x = 2.5$ | $= 0.75$ | cso | A1 |
and $F(x) = 0.25$
$-\frac{1}{9}x^2 + \frac{8}{9}x - \frac{7}{9} = 0.25$
$-x^2 + 8x - 7 = 2.25$ or $-x^2 + 8x - 9.25 = 0$ | quadratic 3 terms = 0 | M1 dep; M1 dep |
$x = \frac{-8 \pm \sqrt{8^2 - 4 \times (-1) \times (-9.25)}}{2 \times (-1)} = 1.40$ | A1 | (6 marks)

## Part (d)
$Q_3 - Q_2 > Q_2 - Q_1$ Or mode = 1 and mode < median Or mean = 2 and median < mode; Sketch of pdf here or be referred to if in a different part of the question; Box plot with $Q_1, Q_2, Q_3$ values marked on; Positive skew | M1, A1 | (2 marks)
A random variable $X$ has probability density function given by

$$\text{f}(x) = \begin{cases} -\frac{2}{9}x + \frac{8}{9} & 1 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}$$

\begin{enumerate}[label=(\alph*)]
\item Show that the cumulative distribution function F(x) can be written in the form $ax^2 + bx + c$, for $1 \leqslant x \leqslant 4$ where $a$, $b$ and $c$ are constants. [3]
\item Define fully the cumulative distribution function F(x). [2]
\item Show that the upper quartile of $X$ is 2.5 and find the lower quartile. [6]
\end{enumerate}

Given that the median of $X$ is 1.88

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item describe the skewness of the distribution. Give a reason for your answer. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2009 Q7 [13]}}
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