| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.8 This is a routine Further Statistics question testing standard probability distribution properties: verifying probabilities sum to 1, calculating E(X) and Var(X) using formulas, and applying linear transformations. All steps are mechanical applications of well-known formulas with straightforward arithmetic. While it's Further Maths content, it requires no problem-solving or insight beyond textbook procedures. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(r\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( \mathrm { X } = \mathrm { r } )\) | \(\frac { 1 } { 35 }\) | \(\frac { 2 } { 35 }\) | \(\frac { 1 } { 7 }\) | \(\frac { 2 } { 7 }\) | \(\frac { 17 } { 35 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | k + 2k + 5k + 10k + 17k = 1 |
| Answer | Marks |
|---|---|
| 35 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 2.4 | |
| 1.1 | AG | |
| 1 | (b) | The distribution has (strong) negative skew |
| [1] | 1.1 | |
| 1 | (c) | 1 29 |
| Answer | Marks |
|---|---|
| ππππππ(ππ) =1 = =1.094β¦ | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | BC Accept any equivalent form. |
| BC Accept any equivalent form. | Decimal answers |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | 245 245 |
| Answer | Marks |
|---|---|
| ππππππ(ππ)=27 = =27.347 | B1FT |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | BC Accept any equivalent form. |
| Answer | Marks |
|---|---|
| FT their Var(X) from (c) | Decimal answers |
Question 1:
1 | (a) | k + 2k + 5k + 10k + 17k = 1
35k = 1 so k = 1
35 | M1
A1
[2] | 2.4
1.1 | AG
1 | (b) | The distribution has (strong) negative skew | B1
[1] | 1.1
1 | (c) | 1 29
πΈπΈ(ππ) =4 = =4.143β¦
7 7
23 268
ππππππ(ππ) =1 = =1.094β¦ | B1
B1
[2] | 1.1a
1.1 | BC Accept any equivalent form.
BC Accept any equivalent form. | Decimal answers
should agree to at least
2 significant figures.
1 | (d) | 245 245
5 75
πΈπΈ(ππ) =10 = =10.714
7 7
17 1340
ππππππ(ππ)=27 = =27.347 | B1FT
B1FT
[2] | 1.1
1.1 | BC Accept any equivalent form.
FT their E(X) from (c)
BC Accept any equivalent form.
FT their Var(X) from (c) | Decimal answers
should agree to at least
2 significant figures.1 The probability distribution of a discrete random variable $X$ is given by the formula $\mathrm { P } ( \mathrm { X } = \mathrm { r } ) = \mathrm { k } \left( ( \mathrm { r } - 1 ) ^ { 2 } + 1 \right)$ for $r = 1,2,3,4,5$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 35 }$.
The distribution of $X$ is shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { r } )$ & $\frac { 1 } { 35 }$ & $\frac { 2 } { 35 }$ & $\frac { 1 } { 7 }$ & $\frac { 2 } { 7 }$ & $\frac { 17 } { 35 }$ \\
\hline
\end{tabular}
\end{center}
\item Comment briefly on the shape of the distribution.
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\end{itemize}
The random variable $Y$ is given by $Y = 5 X - 10$.
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( Y )$
\item $\operatorname { Var } ( Y )$
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2021 Q1 [7]}}