| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a standard S2 piecewise PDF question requiring routine integration for E(X) and Var(X), constructing a CDF with careful handling of the two regions, solving a cubic equation for the median, and comparing mean/median for skewness. While multi-part with several techniques, each step follows textbook procedures without requiring novel insight—slightly easier than average due to the straightforward piecewise structure and standard methods. |
| 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 | Mark | Guidance |
| \(E(X) = \int_0^1 \frac{1}{4}x\, dx + \int_1^2 \frac{x^4}{5}\, dx\) | M1 | For using \(\int xf(x)\,dx\) and adding 2 parts together |
| \(= \left[\frac{x^2}{8}\right]_0^1 + \left[\frac{x^5}{25}\right]_1^2 = \frac{1}{8} + \frac{32}{25} - \frac{1}{25} = \frac{273}{200}\) or \(1.365\) awrt \(\underline{1.37}\) | dM1; A1cso | For all integration correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = \int_0^1 \left(\frac{1}{4}x^2\right)dx + \int_1^2 \left(\frac{1}{5}x^5\right)dx\) | M1 | For using \(\int x^2 f(x)\,dx\) and adding 2 parts |
| \(\text{Var}(X) = \left[\frac{x^3}{12}\right]_0^1 + \left[\frac{x^6}{30}\right]_1^2 - [1.365]^2 = \frac{1}{12} + \frac{64}{30} - \frac{1}{30} - [1.365]^2 = 0.32010..\) \(\underline{0.320}\) | dM1; A1cso | For using \(E(X^2) - [E(X)]^2\); some correct integration must be seen and \(-[E(X)]^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \left(\frac{t^3}{5}\right)dt = \left[\frac{t^4}{20}\right] + D\) and use of \(F(2)=1\) or \(F(1) = \frac{1}{4}\) | M1 | For integrating, using \(+D\) and \(F(2)=1\) or \(F(1)=\frac{1}{4}\) |
| \(F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{4}x & 0 \leqslant x < 1 \\ \frac{x^4}{20} + \frac{1}{5} & 1 \leqslant x \leqslant 2 \\ 1 & x > 2 \end{cases}\) | B1; A1; B1 | B1: correct 2nd row with range; A1: correct 3rd row with range; B1: correct 1st and 4th rows with ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x^4}{20} + \frac{1}{5} = 0.5\) | M1 | If \(F(1) < 0.5\) use 3rd row; if \(F(1) > 0.5\) use 2nd row |
| \(x = \left[\sqrt[4]{6}\right] = 1.565...\) awrt \(\underline{1.57}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean \(<\) median or mode \(= 2\) and mean or median \(<\) mode [or sketch] \(\therefore\) negative skew | M1; A1ft | M1 for correct comparison of mean/median (in \([0,2]\))/mode \(= 2\); A1ft for consistent conclusion about skewness |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \int_0^1 \frac{1}{4}x\, dx + \int_1^2 \frac{x^4}{5}\, dx$ | M1 | For using $\int xf(x)\,dx$ and adding 2 parts together |
| $= \left[\frac{x^2}{8}\right]_0^1 + \left[\frac{x^5}{25}\right]_1^2 = \frac{1}{8} + \frac{32}{25} - \frac{1}{25} = \frac{273}{200}$ or $1.365$ awrt $\underline{1.37}$ | dM1; A1cso | For all integration correct |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = \int_0^1 \left(\frac{1}{4}x^2\right)dx + \int_1^2 \left(\frac{1}{5}x^5\right)dx$ | M1 | For using $\int x^2 f(x)\,dx$ and adding 2 parts |
| $\text{Var}(X) = \left[\frac{x^3}{12}\right]_0^1 + \left[\frac{x^6}{30}\right]_1^2 - [1.365]^2 = \frac{1}{12} + \frac{64}{30} - \frac{1}{30} - [1.365]^2 = 0.32010..$ $\underline{0.320}$ | dM1; A1cso | For using $E(X^2) - [E(X)]^2$; some correct integration must be seen and $-[E(X)]^2$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \left(\frac{t^3}{5}\right)dt = \left[\frac{t^4}{20}\right] + D$ and use of $F(2)=1$ or $F(1) = \frac{1}{4}$ | M1 | For integrating, using $+D$ and $F(2)=1$ or $F(1)=\frac{1}{4}$ |
| $F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{4}x & 0 \leqslant x < 1 \\ \frac{x^4}{20} + \frac{1}{5} & 1 \leqslant x \leqslant 2 \\ 1 & x > 2 \end{cases}$ | B1; A1; B1 | B1: correct 2nd row with range; A1: correct 3rd row with range; B1: correct 1st and 4th rows with ranges |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x^4}{20} + \frac{1}{5} = 0.5$ | M1 | If $F(1) < 0.5$ use 3rd row; if $F(1) > 0.5$ use 2nd row |
| $x = \left[\sqrt[4]{6}\right] = 1.565...$ awrt $\underline{1.57}$ | A1 | |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean $<$ median or mode $= 2$ and mean or median $<$ mode [or sketch] $\therefore$ negative skew | M1; A1ft | M1 for correct comparison of mean/median (in $[0,2]$)/mode $= 2$; A1ft for consistent conclusion about skewness |6. A random variable $X$ has probability density function given by
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 1 } { 4 } & 0 \leqslant x < 1 \\
\frac { x ^ { 3 } } { 5 } & 1 \leqslant x \leqslant 2 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Use algebraic integration to find $\mathrm { E } ( X )$
\item Use algebraic integration to find $\operatorname { Var } ( X )$
\item Define the cumulative distribution function $\mathrm { F } ( x )$ for all values of $x$.
\item Find the median of $X$, giving your answer to 3 significant figures.
\item Comment on the skewness of the distribution, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q6 [14]}}