| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Expected profit or cost problem |
| Difficulty | Moderate -0.8 This is a straightforward discrete probability distribution question requiring basic calculations of expectation and variance for a sum of independent uniform random variables, followed by simple linear transformations. All steps are routine applications of standard formulas with no problem-solving insight needed. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distribution table: \(t = 2,3,4,5,6\) with \(P(T=t) = \frac{1}{9}, \frac{2}{9}, \frac{3}{9}, \frac{2}{9}, \frac{1}{9}\) | M1 | Attempt at distribution of \(T\), or sample space (needs \(\div 9\)). NOT from e.g. \(\frac{6\times2}{3}\) |
| \(E(T) = \frac{1}{9}\times2 + \frac{2}{9}\times3 + \frac{3}{9}\times4 + \frac{2}{9}\times5 + \frac{1}{9}\times6\) | M1 | Attempt sum of \(t \times P(T=t)\) |
| \(= 4\) AG | A1 | Fully correct working ("show that"), www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(S_1) = \frac{1}{3}\times1 + \frac{1}{3}\times2 + \frac{1}{3}\times3\ [=2]\) | M1 | Probabilities for \(S_1\) seen as part of attempt to use \(S_1 + S_2\) (or \(2S_1\)). Or \(\frac{3+1}{2}\times2\) |
| \(E(T) = 2E(S_1)\) | M1 | \(\left(\frac{1}{3}\times1+\frac{1}{3}\times2+\frac{1}{3}\times3\right)^2\) M1M0A0 |
| \(= 4\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(T^2) = 4\times\frac{1}{9}+9\times\frac{2}{9}+16\times\frac{3}{9}+25\times\frac{2}{9}+36\times\frac{1}{9}\ \left[=\frac{156}{9}\right]\) | M1 | Or: \(2^2\times\frac{1}{9}+1^2\times\frac{2}{9}+0^2\times\frac{3}{9}+1^2\times\frac{2}{9}+2^2\times\frac{1}{9}\) M2 |
| \(\text{Var}(T) = E(T^2) - [E(T)]^2 = \frac{156}{9} - 4^2\) | M1 | Allow from no working if \(\frac{3}{3}\) in (a) |
| \(= \frac{4}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(S_1^2) = \frac{1}{3}\times1 + \frac{1}{3}\times4 + \frac{1}{3}\times9 = \frac{14}{3}\) | M1 | Probs for \(S_1\) as part of \(S_1 + S_2\). Or \(\frac{n^2-1}{12}\times2\) |
| \(\text{Var}(S_1) = \frac{14}{3} - 2^2\ \left[=\frac{2}{3}\right]\) | M1 | Not \(2\times\frac{14}{3} - 4^2\) |
| \(\text{Var}(T) = 2\text{Var}(S_1) = \frac{4}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3E(T) - 1.5\ [=(£){-0.3}]\) | B1 | Allow \(-30\) or "\(30p\) loss" etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3^2 \times \text{Var}(T)\) | M1 | \(0.3\times\text{Var}(T)\): M1. FT on their \(\frac{4}{3}\) provided \(> 0\) |
| \(= (£^2)0.12\) or \(\frac{3}{25}\) | A1ft | Not isw if later \(- 1.5\). Ignore units |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distribution table: $t = 2,3,4,5,6$ with $P(T=t) = \frac{1}{9}, \frac{2}{9}, \frac{3}{9}, \frac{2}{9}, \frac{1}{9}$ | M1 | Attempt at distribution of $T$, or sample space (needs $\div 9$). NOT from e.g. $\frac{6\times2}{3}$ |
| $E(T) = \frac{1}{9}\times2 + \frac{2}{9}\times3 + \frac{3}{9}\times4 + \frac{2}{9}\times5 + \frac{1}{9}\times6$ | M1 | Attempt sum of $t \times P(T=t)$ |
| $= 4$ AG | A1 | Fully correct working ("show that"), www |
**Alternative:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(S_1) = \frac{1}{3}\times1 + \frac{1}{3}\times2 + \frac{1}{3}\times3\ [=2]$ | M1 | Probabilities for $S_1$ seen as part of attempt to use $S_1 + S_2$ (or $2S_1$). Or $\frac{3+1}{2}\times2$ |
| $E(T) = 2E(S_1)$ | M1 | $\left(\frac{1}{3}\times1+\frac{1}{3}\times2+\frac{1}{3}\times3\right)^2$ M1M0A0 |
| $= 4$ AG | A1 | |
**[3 marks]**
---
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(T^2) = 4\times\frac{1}{9}+9\times\frac{2}{9}+16\times\frac{3}{9}+25\times\frac{2}{9}+36\times\frac{1}{9}\ \left[=\frac{156}{9}\right]$ | M1 | Or: $2^2\times\frac{1}{9}+1^2\times\frac{2}{9}+0^2\times\frac{3}{9}+1^2\times\frac{2}{9}+2^2\times\frac{1}{9}$ M2 |
| $\text{Var}(T) = E(T^2) - [E(T)]^2 = \frac{156}{9} - 4^2$ | M1 | Allow from no working if $\frac{3}{3}$ in (a) |
| $= \frac{4}{3}$ | A1 | |
**Alternative:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(S_1^2) = \frac{1}{3}\times1 + \frac{1}{3}\times4 + \frac{1}{3}\times9 = \frac{14}{3}$ | M1 | Probs for $S_1$ as part of $S_1 + S_2$. Or $\frac{n^2-1}{12}\times2$ |
| $\text{Var}(S_1) = \frac{14}{3} - 2^2\ \left[=\frac{2}{3}\right]$ | M1 | Not $2\times\frac{14}{3} - 4^2$ |
| $\text{Var}(T) = 2\text{Var}(S_1) = \frac{4}{3}$ | A1 | |
**[3 marks]**
---
## Part (c)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3E(T) - 1.5\ [=(£){-0.3}]$ | B1 | Allow $-30$ or "$30p$ loss" etc |
**[1 mark]**
---
## Part (c)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3^2 \times \text{Var}(T)$ | M1 | $0.3\times\text{Var}(T)$: M1. FT on their $\frac{4}{3}$ provided $> 0$ |
| $= (£^2)0.12$ or $\frac{3}{25}$ | A1ft | Not isw if later $- 1.5$. Ignore units |
**[2 marks]**1 When a spinner is spun, the outcome is equally likely to be 1,2 or 3 . In a competition, the spinner is spun twice and the outcomes are added to give a total score $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that the expectation of $T$ is 4 .
\item Find the variance of $T$.
A competitor pays $\pounds 1.50$ to enter the competition and receives $\pounds X$, where $X = 0.3 T$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the expectation of the competitor's profit.
\item Find the variance of the competitor's profit.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2019 Q1 [9]}}