| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from independent trials |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring basic calculations with given probabilities. Part (a) is simple verification using P(X=4) = 0.6^4. Part (b) involves standard E(X) and Var(X) calculations from a discrete distribution table. Part (c) requires conceptual understanding of how transformations affect expectation and variance, but no actual computation. While it tests multiple concepts, all are routine applications of A-level statistics formulas with no novel problem-solving required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(r\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | 0.4 | 0.24 | 0.144 | 0.0864 | 0.1296 |
|
| ||||
| 0 or 1 | 0 | ||||
| 2 | 2 | ||||
| 3 | 3 | ||||
| 4 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 0.64 = 0.1296 |
| [1] | 1.1 | Given answer |
| 1 | (b) | E(X) = 1.3(056…) |
| Var(X) = 1.96(3…) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 1.1 | Accept 2.0 | |
| 1 | (c) | E(Y) will be less |
| Answer | Marks |
|---|---|
| more spread out. | E1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | Need to refer to change from X to Y |
Question 1:
1 | (a) | 0.64 = 0.1296 | B1
[1] | 1.1 | Given answer
1 | (b) | E(X) = 1.3(056…)
Var(X) = 1.96(3…) | B1
B1
[2] | 1.1
1.1 | Accept 2.0
1 | (c) | E(Y) will be less
since the value of r = 1 for X becomes r = 0 for Y
and r = 4 becomes r = 5 but P(X = 1) > P(X = 4).
Var(Y) will greater since the distribution is now
more spread out. | E1
E1
E1
[3] | 3.1a
2.1
2.2a | Need to refer to change from X to Y
Allow other suitable answers1 In a quiz a contestant is asked up to four questions. The contestant's turn ends once the contestant gets a question wrong or has answered all four questions. The probability that a particular contestant gets any question correct is 0.6 , independently of other questions. The discrete random variable $X$ models the number of questions which the contestant gets correct in a turn.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 4 ) = 0.1296$.
The probability distribution of $X$ is shown in Fig. 1.1.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & 0.4 & 0.24 & 0.144 & 0.0864 & 0.1296 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{table}
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\end{itemize}
The number of points that a contestant scores is as shown in Fig. 1.2.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | }
\hline
\begin{tabular}{ c }
Number of \\
questions correct \\
\end{tabular} & \begin{tabular}{ c }
Number of \\
points scored \\
\end{tabular} \\
\hline
0 or 1 & 0 \\
\hline
2 & 2 \\
\hline
3 & 3 \\
\hline
4 & 5 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{table}
The discrete random variable $Y$ models the number of points which the contestant scores.
\item Without doing any working, explain whether each of the following will be less than, equal to or greater than the corresponding value for $X$.
\begin{itemize}
\item $\mathrm { E } ( Y )$
\item $\operatorname { Var } ( Y )$
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2022 Q1 [6]}}