| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Cumulative distribution function |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing basic properties of continuous uniform distributions. All parts involve direct application of standard formulas: reading from CDF, recognizing uniform distribution, using E(X)=(a+b)/2 and Var(X)=(b-a)²/12. No problem-solving or novel insight required, just routine recall and calculation. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X>4) = 1 - F(4)\) | M1 | Writing or using \(1-F(4)\) oe |
| \(= 1 - \frac{3}{5}\) | ||
| \(= \frac{2}{5}\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = \frac{dF(x)}{dx} = \frac{1}{5}\) | M1 | M1 for differentiating to get \(\frac{1}{5}\) |
| \(f(x) = \begin{cases} \frac{1}{5} & 1 \leq x \leq 6 \\ 0 & \text{otherwise} \end{cases}\) | A1 | A1 both lines correct with ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 3.5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Variance} = \frac{(6-1)^2}{12}\) or \(\int_1^6 \frac{1}{5}x^2\,dx - (3.5)^2\) | M1 | M1 \(\frac{(6-1)^2}{12}\) or \(\int_1^6\frac{1}{5}x^2\,dx\) − 'their \(3.5\)'\(^2\) |
| \(= \frac{25}{12}\) awrt \(2.08\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X^2) = \text{Var}(X) + [E(X)]^2 = \frac{25}{12} + 3.5^2\) or \(\int_1^6\frac{1}{5}x^2\,dx\) or \(\int_1^6\frac{1}{5}(3x^2+1)\,dx\) | M1 | 1st M1 'their \(\text{Var}(X)\)' + ['their \(E(X)\)']\(^2\) (must follow from 1st method in (e)) or \(\int_1^6\frac{1}{5}x^2\,dx\) and integrating \(x^n \to \frac{x^{n+1}}{n+1}\) or writing \(\int_1^6\frac{1}{5}(3x^2+1)\,dx\) |
| \(= \frac{43}{3}\) | May be implied by \(\frac{43}{3}\) seen | |
| \(E(3X^2+1) = 3E(X^2)+1\) | dM1 | 2nd M1 (dependent) using \(3 \times\) 'their \(E(X^2)\)' \(+ 1\) or \(\int_1^6\frac{1}{5}(3x^2+1)\,dx\) and integrating \(x^n \to \frac{x^{n+1}}{n+1}\) |
| \(= \left[\frac{3x^3}{15}+\frac{x}{5}\right]_1^6 = 44\) | ||
| \(= 44\) | A1cao |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X>4) = 1 - F(4)$ | M1 | Writing or using $1-F(4)$ oe |
| $= 1 - \frac{3}{5}$ | | |
| $= \frac{2}{5}$ oe | A1 | |
**(2 marks)**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1$ | B1 | |
**(1 mark)**
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = \frac{dF(x)}{dx} = \frac{1}{5}$ | M1 | M1 for differentiating to get $\frac{1}{5}$ |
| $f(x) = \begin{cases} \frac{1}{5} & 1 \leq x \leq 6 \\ 0 & \text{otherwise} \end{cases}$ | A1 | A1 both lines correct with ranges |
**(2 marks)**
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 3.5$ | B1 | |
**(1 mark)**
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Variance} = \frac{(6-1)^2}{12}$ or $\int_1^6 \frac{1}{5}x^2\,dx - (3.5)^2$ | M1 | M1 $\frac{(6-1)^2}{12}$ or $\int_1^6\frac{1}{5}x^2\,dx$ − 'their $3.5$'$^2$ |
| $= \frac{25}{12}$ awrt $2.08$ | A1 | |
**(2 marks)**
## Part (f)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^2) = \text{Var}(X) + [E(X)]^2 = \frac{25}{12} + 3.5^2$ or $\int_1^6\frac{1}{5}x^2\,dx$ or $\int_1^6\frac{1}{5}(3x^2+1)\,dx$ | M1 | 1st M1 'their $\text{Var}(X)$' + ['their $E(X)$']$^2$ (must follow from 1st method in (e)) **or** $\int_1^6\frac{1}{5}x^2\,dx$ and integrating $x^n \to \frac{x^{n+1}}{n+1}$ **or** writing $\int_1^6\frac{1}{5}(3x^2+1)\,dx$ |
| $= \frac{43}{3}$ | | May be implied by $\frac{43}{3}$ seen |
| $E(3X^2+1) = 3E(X^2)+1$ | dM1 | 2nd M1 (dependent) **using** $3 \times$ 'their $E(X^2)$' $+ 1$ **or** $\int_1^6\frac{1}{5}(3x^2+1)\,dx$ and integrating $x^n \to \frac{x^{n+1}}{n+1}$ |
| $= \left[\frac{3x^3}{15}+\frac{x}{5}\right]_1^6 = 44$ | | |
| $= 44$ | A1cao | |
**(3 marks)**2. A continuous random variable $X$ has cumulative distribution function
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x < 1 \\
\frac { 1 } { 5 } ( x - 1 ) & 1 \leqslant x \leqslant 6 \\
1 & x > 6
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 4 )$
\item Write down the value of $\mathrm { P } ( X \neq 4 )$
\item Find the probability density function of $X$, specifying it for all values of $X$
\item Write down the value of $\mathrm { E } ( X )$
\item Find $\operatorname { Var } ( X )$
\item Hence or otherwise find $\mathrm { E } \left( 3 X ^ { 2 } + 1 \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2015 Q2 [11]}}