| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(aX+b) or Var(aX+b) given distribution |
| Difficulty | Moderate -0.8 This is a straightforward application of standard expectation and variance formulas with a discrete probability distribution, plus basic linear transformations (E(aX+b) and Var(aX+b)). The distribution is symmetric making E(X)=2 obvious by inspection. The final parts on sampling are simple recall of basic statistical concepts. All parts are routine textbook exercises requiring no problem-solving insight. |
| Spec | 2.01a Population and sample: terminology2.01c Sampling techniques: simple random, opportunity, etc5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
| \(P(X = x)\) | \(0.05\) | \(0.2\) | \(0.5\) | \(0.2\) | \(0.05\) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | (A) |
| 2 | B1 | |
| [1] | M | |
| 2.4 | E |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | (B) |
| Answer | Marks |
|---|---|
| = 0.8 | I |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | For Σr2p (at least 3 terms |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (ii) | P |
| Answer | Marks |
|---|---|
| Var(Y) = 2502 × 0.8 = 50000 | B1FT |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | FT their E(X) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (iii) | Testing all of them not suitable as they will not be |
| available to sell to customers. | E1 | |
| [1] | 2.4 | OR It would take too long |
| 2 | (iv) | Random sample avoids unsuspected sources of bias. |
| [1] | 1.2 | OR allows inference |
Question 2:
2 | (i) | (A) | The distribution is symmetrical so E(X) is at the centre i.e.
2 | B1
[1] | M
2.4 | E
Or use of
0(cid:117)0.05(cid:14)1(cid:117)0.2(cid:14)2(cid:117)0.5...
(cid:14)3(cid:117)0.2(cid:14)4(cid:117)0.05
2 | (i) | (B) | E(X 2) = (02 × 0.05) + (12 × 0.2) + (22 × 0.5) + (32 × 0.2) +
(42 × 0.05)
E
= 4.8
Var (X) = 4.8 – 22
= 0.8 | I
C
M1
M1
A1FT
[3] | 1.1
1.2
1.1 | For Σr2p (at least 3 terms
correct)
dep for – their E(X)²
FT their E(X) provided Var(X)
> 0
2 | (ii) | P
S
E(Y) = 250 × 2 – 80 = 420
Var(Y) = 2502 × 0.8 = 50000 | B1FT
B1FT
[2] | 1.1
1.1 | FT their E(X)
FT their Var(X)
2 | (iii) | Testing all of them not suitable as they will not be
available to sell to customers. | E1
[1] | 2.4 | OR It would take too long
2 | (iv) | Random sample avoids unsuspected sources of bias. | E1
[1] | 1.2 | OR allows inferenceThe number of televisions of a particular model sold per week at a retail store can be modelled by a random variable $X$ with the probability function shown in the table.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & $0$ & $1$ & $2$ & $3$ & $4$ \\
\hline
$P(X = x)$ & $0.05$ & $0.2$ & $0.5$ & $0.2$ & $0.05$ \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Explain why $\text{E}(X) = 2$. [1]
\item Find $\text{Var}(X)$. [3]
\end{enumerate}
\item The profit, measured in pounds made in a week, on the sales of this model of television is given by $Y$, where $Y = 250X - 80$.
Find
\begin{itemize}
\item $\text{E}(Y)$ and
\item $\text{Var}(Y)$. [2]
\end{itemize}
\end{enumerate}
The remote controls for the televisions are quality tested by the manufacturer to see how long they last before they fail.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Explain why it would be inappropriate to test all the remote controls in this way. [1]
\item State an advantage of using random sampling in this context. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor Q2 [8]}}