| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with k |
| Difficulty | Standard +0.3 This is a standard S2 piecewise PDF question requiring: (a) finding k by setting the PDF continuous and using the integral=1 condition (routine calculus), (b) identifying the mode by inspection of where the PDF is maximum, and (c) applying conditional probability with integration. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^k \left(\frac{2x}{15}\right)dx + \int_5^k \frac{1}{5}(5-x)dx = 1\) | M1 | Complete method with correct limits, set equal to 1 |
| \(\left[\frac{x^2}{15}\right]_0^k + \left[x - \frac{x^2}{10}\right]_k^5 = 1\) | M1 | Evidence of \(x^n \to x^{n+1}\) |
| Both \(\frac{2x}{15} \to \frac{x^2}{15}\) and \(\frac{1}{5}(5-x) \to x - \frac{x^2}{10}\) | A1 o.e. | |
| \(\left(\frac{k^2}{15}\right) + \left(5 - \frac{5^2}{10} - \left(k - \frac{k^2}{10}\right)\right) = 1\) | ||
| \(2k^2 + 150 - 75 - 30k + 3k^2 = 30\) | ||
| \(k^2 - 6k + 9 = 0\) or \(\frac{k^2}{6} - k + \frac{3}{2} = 0\) | ||
| \((k-3)(k-3) = 0 \Rightarrow k = \ldots\) | dM1 | Dependent on 1st M; attempt to solve 3-term quadratic leading to \(k=\) |
| \(k = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\text{mode} =\}\ 3\) | B1 ft | 3 or states their \(k\) value from part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P\left(X \leqslant \frac{k}{2} \,\middle | \, X \leqslant k\right) = \frac{P\left(X \leqslant \frac{k}{2} \cap X \leqslant k\right)}{P(X \leqslant k)}\) | |
| \(= \frac{P\left(X \leqslant \frac{k}{2}\right)}{P(X \leqslant k)}\) | M1 | Either \(\frac{P\left(X \leqslant \frac{k}{2}\right)}{P(X \leqslant k)}\) or \(\frac{F\!\left(\frac{k}{2}\right)}{F(k)}\) seen or implied |
| \(= \dfrac{\displaystyle\int_0^{k/2} \frac{2x}{15}\,dx}{\displaystyle\int_0^k \frac{2x}{15}\,dx}\) | dM1 | Dependent on 1st M; applies conditional probability with correct integrals |
| \(= \dfrac{\frac{1}{15}\!\left(\frac{k}{2}\right)^2}{\dfrac{k^2}{15}}\) | A1 ft | Correct substitution of limits or \(k\) into conditional probability formula |
| \(\left\{= \frac{\left(\frac{9}{60}\right)}{\left(\frac{9}{15}\right)} = \frac{0.15}{0.6}\right\} = \frac{1}{4}\) | A1 cao | \(\frac{1}{4}\) or \(0.25\) |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^k \left(\frac{2x}{15}\right)dx + \int_5^k \frac{1}{5}(5-x)dx = 1$ | M1 | Complete method with correct limits, set equal to 1 |
| $\left[\frac{x^2}{15}\right]_0^k + \left[x - \frac{x^2}{10}\right]_k^5 = 1$ | M1 | Evidence of $x^n \to x^{n+1}$ |
| Both $\frac{2x}{15} \to \frac{x^2}{15}$ and $\frac{1}{5}(5-x) \to x - \frac{x^2}{10}$ | A1 o.e. | |
| $\left(\frac{k^2}{15}\right) + \left(5 - \frac{5^2}{10} - \left(k - \frac{k^2}{10}\right)\right) = 1$ | | |
| $2k^2 + 150 - 75 - 30k + 3k^2 = 30$ | | |
| $k^2 - 6k + 9 = 0$ or $\frac{k^2}{6} - k + \frac{3}{2} = 0$ | | |
| $(k-3)(k-3) = 0 \Rightarrow k = \ldots$ | dM1 | Dependent on 1st M; attempt to solve 3-term quadratic leading to $k=$ |
| $k = 3$ | A1 | |
**[5 marks]**
---
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{\text{mode} =\}\ 3$ | B1 ft | 3 or states their $k$ value from part (a) |
**[1 mark]**
---
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P\left(X \leqslant \frac{k}{2} \,\middle|\, X \leqslant k\right) = \frac{P\left(X \leqslant \frac{k}{2} \cap X \leqslant k\right)}{P(X \leqslant k)}$ | | |
| $= \frac{P\left(X \leqslant \frac{k}{2}\right)}{P(X \leqslant k)}$ | M1 | Either $\frac{P\left(X \leqslant \frac{k}{2}\right)}{P(X \leqslant k)}$ or $\frac{F\!\left(\frac{k}{2}\right)}{F(k)}$ seen or implied |
| $= \dfrac{\displaystyle\int_0^{k/2} \frac{2x}{15}\,dx}{\displaystyle\int_0^k \frac{2x}{15}\,dx}$ | dM1 | Dependent on 1st M; applies conditional probability with correct integrals |
| $= \dfrac{\frac{1}{15}\!\left(\frac{k}{2}\right)^2}{\dfrac{k^2}{15}}$ | A1 ft | Correct substitution of limits or $k$ into conditional probability formula |
| $\left\{= \frac{\left(\frac{9}{60}\right)}{\left(\frac{9}{15}\right)} = \frac{0.15}{0.6}\right\} = \frac{1}{4}$ | A1 cao | $\frac{1}{4}$ or $0.25$ |
**[4 marks]**
**Total: [10 marks]**\begin{enumerate}
\item A random variable $X$ has probability density function
\end{enumerate}
$$f ( x ) = \begin{cases} \frac { 2 x } { 15 } & 0 \leqslant x \leqslant k \\ \frac { 1 } { 5 } ( 5 - x ) & k < x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
(a) Showing your working clearly, find the value of $k$.\\
(b) Write down the mode of $X$.\\
(c) Find $\mathrm { P } \left( \left. X \leqslant \frac { k } { 2 } \right\rvert \, X \leqslant k \right)$\\
\hfill \mbox{\textit{Edexcel S2 2015 Q7 [10]}}