Edexcel S2 2010 January — Question 2 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2010
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeCumulative distribution function
DifficultyModerate -0.8 This is a straightforward S2 question testing basic understanding of CDFs and uniform distributions. All parts require direct application of standard definitions: reading values from the CDF, differentiating to find the PDF, recognizing a uniform distribution, and applying standard formulas for mean and variance. No problem-solving or novel insight required—purely procedural recall.
Spec5.02e Discrete uniform distribution5.03a Continuous random variables: pdf and cdf5.03d E(g(X)): general expectation formula5.03e Find cdf: by integration

2. A continuous random variable \(X\) has cumulative distribution function $$\mathrm { F } ( x ) = \begin{cases} 0 , & x < - 2 \\ \frac { x + 2 } { 6 } , & - 2 \leqslant x \leqslant 4 \\ 1 , & x > 4 \end{cases}$$
  1. Find \(\mathrm { P } ( X < 0 )\).
  2. Find the probability density function \(\mathrm { f } ( x )\) of \(X\).
  3. Write down the name of the distribution of \(X\).
  4. Find the mean and the variance of \(X\).
  5. Write down the value of \(\mathrm { P } ( X = 1 )\).

AnswerMarks Guidance
(a) \(P(X < 0) = F(0) = \frac{2}{6} = \frac{1}{3}\)M1 A1 (2 marks)
(b) \(f(x) = \frac{dF(x)}{dx}\) where \(f(x) = \begin{cases} \frac{1}{6} & -2 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}\)M1 A1 B1 (3 marks)
(c) Continuous Uniform (Rectangular) distributionB1 (1 mark)
(d) Mean \(= 1\)B1 Variance \(= \frac{(4-(-2))^2}{12} = 3\)
(e) \(P(X = 1) = 0\)B1 (1 mark)
Total [10]
**(a)** $P(X < 0) = F(0) = \frac{2}{6} = \frac{1}{3}$ | M1 A1 | (2 marks)

**(b)** $f(x) = \frac{dF(x)}{dx}$ where $f(x) = \begin{cases} \frac{1}{6} & -2 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$ | M1 A1 B1 | (3 marks)

**(c)** Continuous Uniform (Rectangular) distribution | B1 | (1 mark)

**(d)** Mean $= 1$ | B1 | Variance $= \frac{(4-(-2))^2}{12} = 3$ | M1 A1 | (3 marks)

**(e)** $P(X = 1) = 0$ | B1 | (1 mark)

| **Total [10]**

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2. A continuous random variable $X$ has cumulative distribution function

$$\mathrm { F } ( x ) = \begin{cases} 0 , & x < - 2 \\ \frac { x + 2 } { 6 } , & - 2 \leqslant x \leqslant 4 \\ 1 , & x > 4 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X < 0 )$.
\item Find the probability density function $\mathrm { f } ( x )$ of $X$.
\item Write down the name of the distribution of $X$.
\item Find the mean and the variance of $X$.
\item Write down the value of $\mathrm { P } ( X = 1 )$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2010 Q2 [10]}}