| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Complete probability distribution table |
| Difficulty | Moderate -0.3 This is a straightforward hypergeometric distribution question requiring standard calculations. Part (a) is a simple 'show that' using combinations, part (b) involves routine expectation and variance calculations from a given distribution table, and part (c) requires basic reasoning about conditional probability effects. While it involves multiple parts and some calculation, it requires no novel insight and is typical textbook material for Further Statistics. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(r\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 1 } { 99 }\) | \(\frac { 14 } { 99 }\) | \(\frac { 42 } { 99 }\) | \(\frac { 35 } { 99 }\) | \(\frac { 7 } { 99 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | P(X = 4) = |
| Answer | Marks | Guidance |
|---|---|---|
| 12×11×10×9 = 99 | B1 | |
| [1] | 1.1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (b) | E(X) = or 2.33(3…) |
| Answer | Marks |
|---|---|
| 99 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | BC |
| BC | ft their E(X) if working |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (c) | P(X = 4) is higher than P(X = 0) and the other |
| Answer | Marks |
|---|---|
| E(X) would be smaller because… | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| 3.1a | Consideration of both (may then |
| Answer | Marks |
|---|---|
| attempt to explain | NB if calculated: |
Question 1:
1 | (a) | P(X = 4) =
7 6 5 4 7
12×11×10×9 = 99 | B1
[1] | 1.1 | AG | or: C ÷ C
7 4 12 4
7
=99
1 | (b) | E(X) = or 2.33(3…)
7
3
Var(X) = or 0.707(07…) or 0.71
70
99 | B1
B1ft
[2] | 1.1a
1.1 | BC
BC | ft their E(X) if working
shown
1 | (c) | P(X = 4) is higher than P(X = 0) and the other
probabilities would increase in proportion.
E(X) would be smaller because… | E1
B1
[2] | 2.4
3.1a | Consideration of both (may then
come to wrong conclusion)
Correct decision linked to (a partial)
attempt to explain | NB if calculated:
E(girls out of 2 chosen)
= 1.2 then
E(X) = 2.2 with
conclusion1 A quiz team of 4 students is to be selected from a group of 7 girls and 5 boys. The team is selected at random from the students in the group. The number of girls in the team is denoted by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 4 ) = \frac { 7 } { 99 }$.
Table 1 shows the probability distribution of $X$.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 99 }$ & $\frac { 14 } { 99 }$ & $\frac { 42 } { 99 }$ & $\frac { 35 } { 99 }$ & $\frac { 7 } { 99 }$ \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\end{itemize}
It is decided that the quiz team must have at least 1 girl and at least 1 boy, but the team is still otherwise selected at random.
\item Explain whether $\mathrm { E } ( X )$ would be smaller than, equal to or larger than the value which you found in part (b).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2020 Q1 [5]}}