| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.3 This is a straightforward discrete probability distribution question requiring routine calculations: substituting into a given formula, using ΣP(X=r)=1 to find k, computing E(X) and Var(X) using standard formulas, and solving a simple linear equation for expected profit. All techniques are standard AS-level statistics with no novel problem-solving required, making it slightly easier than average. |
| 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 |
| \(\mathrm { P } ( X = r )\) | \(7 k\) | \(3 k\) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | r |
| [1] | 1.1 | |
| P(X = r) | 9k | 7k |
| 1 | (b) | 9k + 7k + 5k + 3k + k = 1 ⇒ k = 1 o.e. |
| 25 | B1 | |
| [1] | 2.2a | |
| 1 | (c) | E(X)=11 (= 2.2) |
| Answer | Marks |
|---|---|
| V a r ( X ) = 32 45 (=1.36) | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | BC FT 55×their k |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | E(Winnings) = 50 × 2.2 = 110 |
| C = 110 + 25 = 135 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | FT 50 × their E(X) for M1 soi |
Question 1:
1 | (a) | r | 1 | 2 | 3 | 4 | 5 | B1
[1] | 1.1
P(X = r) | 9k | 7k | 5k | 3k | k
1 | (b) | 9k + 7k + 5k + 3k + k = 1 ⇒ k = 1 o.e.
25 | B1
[1] | 2.2a
1 | (c) | E(X)=11 (= 2.2)
5
V a r ( X ) = 32 45 (=1.36) | B1
B1
[2] | 1.1a
1.1 | BC FT 55×their k
BC FT 155×their k – their [E(X)]2 provided Var(X) > 0
1 | (d) | E(Winnings) = 50 × 2.2 = 110
C = 110 + 25 = 135 | M1
A1
[2] | 1.1
1.1 | FT 50 × their E(X) for M1 soi
FT their E(X)
Condone £1.35 and 135p1 A fair five-sided spinner has sectors labelled 1, 2, 3, 4, 5. In a game at a stall at a charity event, the spinner is spun twice. The random variable $X$ represents the lower of the two scores. The probability distribution of $X$ is given by the formula\\
$\mathrm { P } ( \mathrm { X } = \mathrm { r } ) = \mathrm { k } ( 11 - 2 \mathrm { r } )$ for $r = 1,2,3,4,5$,\\
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Complete the copy of this table in the Printed Answer Booklet.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = r )$ & & $7 k$ & & $3 k$ & \\
\hline
\end{tabular}
\end{center}
\item Determine the value of $k$.
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\item The stall-holder charges a player $C$ pence to play the game, and then pays the player $50 X$ pence, where $X$ is the player's score.
\end{itemize}
Given that the average profit that the stall-holder makes on one game is 25 pence, find the value of $C$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2022 Q1 [6]}}