| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Expectation of function of X |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard integration techniques for E(X), Var(X), and probability calculations with a given pdf, followed by basic expectation calculations using conditional values. All steps are routine applications of formulas with no conceptual challenges beyond A-level S2 curriculum. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^2 \frac{9x^2}{10} \cdot \frac{3x^3}{10}\,dx = \left[\frac{3x^3}{10} - \frac{3x^4}{40}\right]_0^2\) | M1A1 | M1: using \(\int xf(x)\) and attempting to integrate. At least one \(x^n \to x^{n+1}\). Ignore limits. A1: Correct integration, ignore limits |
| \(= \left(\frac{3\times2^3}{10} - \frac{3\times2^4}{40}\right)\) | M1d | M1d: substituting correct limits, dependent on previous Method mark being awarded |
| \(= 1.2\) | A1 (4) | A1: 1.2 oe. Allow 1.20 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = \int_0^2 \frac{9x^3}{10} - \frac{3x^4}{10}\,dx = \left[\frac{9x^4}{40} - \frac{3x^5}{50}\right]_0^2\) | M1 | M1: using \(\int x^2 f(x)\) and attempting to integrate. At least one \(x^n \to x^{n+1}\). Ignore limits |
| \(= \frac{42}{25} = 1.68\) | A1 | A1: Allow equivalent fractions. May be implied by correct answer. Condone \(\text{Var}(X) = 1.68\) |
| \(\text{Var}(X) = 1.68 - 1.2^2\) | M1d | M1d: use of \(E(X^2) - E(X)^2\) |
| \(= 0.24\) | A1 (4) | A1: cao, allow 0.240 or 6/25 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([P(X>1.5)=]\ \int_{1.5}^{2}\frac{9x}{10}-\frac{3x^2}{10}\,dx\) or \(1-\int_0^{1.5}\frac{9x}{10}-\frac{3x^2}{10}\,dx\) | M1 | M1: writing or using correct integral with correct limits, or using \(1-F(1.5)\) for this distribution |
| \(=\left[\frac{9x^2}{20}-\frac{3x^3}{30}\right]_{1.5}^{2}\) or \(1-\left[\frac{9x^2}{20}-\frac{3x^3}{30}\right]_0^{1.5}\) | A1 | A1: Correct integration. Condone missing 1 |
| \(= \frac{13}{40} = 0.325\) | A1cso (3) | A1cso: 0.325 or 13/40 oe |
| NB: Watch out for using \(1-f(1.5)\) or \(1-\frac{9(1.5)-3(1.5)^2}{10}\) — This gets M0A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0.325)\times25 + (1-0.325)\times50 = \text{£}41.875\) | M1A1 (2) | M1: \((their(c))\times25+(1-their(c))\times50\). Allow use of their part (c) or 0.325. Allow \(50-(part(c))\times25\). A1: awrt 41.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| £\(50\times0.8\) or £40 or 0.4 or awrt 0.038 or awrt 0.163 | M1 | M1: Allow \((50\times0.8)n\) or £\(40n\ (n\neq0)\). NB Allow 20% off (of) 50 = £40 |
| Peter should not remove the staples as the expected amount earned per bin will be less. | A1ft (2) | A1ft: Correct statement containing the word staples and one of the 4 comparisons (ft on (c) or (d)) or the difference in these values must be seen. £\(40n <\) part(d)\(\times n\), or \(0.4 <\) their part (c) or \(0.6 < 1-\)their part(c), or awrt \(0.838 > 0.8\) or \(0.162 < 0.2\) |
| Total 15 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^2 \frac{9x^2}{10} \cdot \frac{3x^3}{10}\,dx = \left[\frac{3x^3}{10} - \frac{3x^4}{40}\right]_0^2$ | M1A1 | M1: using $\int xf(x)$ and attempting to integrate. At least one $x^n \to x^{n+1}$. Ignore limits. A1: Correct integration, ignore limits |
| $= \left(\frac{3\times2^3}{10} - \frac{3\times2^4}{40}\right)$ | M1d | M1d: substituting correct limits, dependent on previous Method mark being awarded |
| $= 1.2$ | A1 **(4)** | A1: 1.2 oe. Allow 1.20 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = \int_0^2 \frac{9x^3}{10} - \frac{3x^4}{10}\,dx = \left[\frac{9x^4}{40} - \frac{3x^5}{50}\right]_0^2$ | M1 | M1: using $\int x^2 f(x)$ and attempting to integrate. At least one $x^n \to x^{n+1}$. Ignore limits |
| $= \frac{42}{25} = 1.68$ | A1 | A1: Allow equivalent fractions. May be implied by correct answer. Condone $\text{Var}(X) = 1.68$ |
| $\text{Var}(X) = 1.68 - 1.2^2$ | M1d | M1d: use of $E(X^2) - E(X)^2$ |
| $= 0.24$ | A1 **(4)** | A1: cao, allow 0.240 or 6/25 oe |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P(X>1.5)=]\ \int_{1.5}^{2}\frac{9x}{10}-\frac{3x^2}{10}\,dx$ or $1-\int_0^{1.5}\frac{9x}{10}-\frac{3x^2}{10}\,dx$ | M1 | M1: writing or using correct integral with correct limits, or using $1-F(1.5)$ for this distribution |
| $=\left[\frac{9x^2}{20}-\frac{3x^3}{30}\right]_{1.5}^{2}$ or $1-\left[\frac{9x^2}{20}-\frac{3x^3}{30}\right]_0^{1.5}$ | A1 | A1: Correct integration. Condone missing 1 |
| $= \frac{13}{40} = 0.325$ | A1cso **(3)** | A1cso: 0.325 or 13/40 oe |
| **NB:** Watch out for using $1-f(1.5)$ or $1-\frac{9(1.5)-3(1.5)^2}{10}$ — **This gets M0A0A0** | | |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0.325)\times25 + (1-0.325)\times50 = \text{£}41.875$ | M1A1 **(2)** | M1: $(their(c))\times25+(1-their(c))\times50$. Allow use of their part (c) or 0.325. Allow $50-(part(c))\times25$. A1: awrt 41.9 |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| £$50\times0.8$ or £40 or 0.4 or awrt 0.038 or awrt 0.163 | M1 | M1: Allow $(50\times0.8)n$ or £$40n\ (n\neq0)$. NB Allow 20% off (of) 50 = £40 |
| Peter should not remove the **staples** as the expected amount earned per bin will be less. | A1ft **(2)** | A1ft: Correct statement containing the word **staples** and one of the 4 comparisons (ft on (c) or (d)) or the difference in these values must be seen. £$40n <$ part(d)$\times n$, or $0.4 <$ their part (c) or $0.6 < 1-$their part(c), or awrt $0.838 > 0.8$ or $0.162 < 0.2$ |
| | **Total 15** | |7. The weight, $X \mathrm {~kg}$, of staples in a bin full of paper has probability density function
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 9 x - 3 x ^ { 2 } } { 10 } & 0 \leqslant x < 2 \\
0 & \text { otherwise }
\end{array} \right.$$
Use integration to find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\item $\mathrm { P } ( X > 1.5 )$
Peter raises money by collecting paper and selling it for recycling. A bin full of paper is sold for $\pounds 50$ but if the weight of the staples exceeds 1.5 kg it sells for $\pounds 25$
\item Find the expected amount of money Peter raises per bin full of paper.
Peter could remove all the staples before the paper is sold but the time taken to remove the staples means that Peter will have $20 \%$ fewer bins full of paper to sell.
\item Decide whether or not Peter should remove all the staples before selling the bins full of paper. Give a reason for your answer.\\
\href{http://PhysicsAndMathsTutor.com}{PhysicsAndMathsTutor.com}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q7 [15]}}