| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Conditional or compound probability scenarios |
| Difficulty | Standard +0.3 This is a straightforward question testing basic discrete uniform distribution and geometric distribution. Parts (a)-(c) require simple recall and standard formula application. Part (d) involves calculating expected value using linearity of expectation with a geometric distribution, which is a standard technique. The question is slightly easier than average as it guides students through each step clearly with no novel problem-solving required. |
| Spec | 5.02e Discrete uniform distribution5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | (Discrete) Uniform or U... |
| on the values 1, 2, …, 32 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 2.5 | Correct distribution stated. |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | 1/32 or 0.03125 or 3.12510–2 |
| [1] | 1.1 | Exact value seen. |
| 2 | (c) | Y ~ Geo(1/32) |
| Answer | Marks |
|---|---|
| = awrt 0.379 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Correct geometric distribution used. |
| Answer | Marks |
|---|---|
| awrt 0.379 | ( 1) 4 ( 1) 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (d) | E(Y) = 32 soi |
| Answer | Marks |
|---|---|
| = –16 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Using E(Y) = 1/p for Geometric |
| Answer | Marks |
|---|---|
| If M0 then SCB1 for –16 | or expected score per turn = |
Question 2:
2 | (a) | (Discrete) Uniform or U...
on the values 1, 2, …, 32 | M1
A1
[2] | 1.1a
2.5 | Correct distribution stated.
Condone “1 to 32”, or equivalent,
provided it is clear that this is a
discrete distribution.
2 | (b) | 1/32 or 0.03125 or 3.12510–2 | B1
[1] | 1.1 | Exact value seen.
2 | (c) | Y ~ Geo(1/32)
P(Y 15) = 1 – (31/32)15
= awrt 0.379 | M1
M1
A1
[3] | 1.1a
1.1
1.1 | Correct geometric distribution used.
soi.
Any correct formulation of the
required probability seen.
If M1M1 not awarded, SCB1 for
awrt 0.379 | ( 1) 4 ( 1) 3
eg 33 12 13 + 33 12 13 + + 13
2 2 2
2 | (d) | E(Y) = 32 soi
So expected final score = –1 31 + 15
= –16 | B1
M1
A1
[3] | 1.1
3.1b
1.1 | Using E(Y) = 1/p for Geometric
Distribution.
For –1 (“32” – 1) + 15. Method
seen.
If M0 then SCB1 for –16 | or expected score per turn =
–1(31/32) + 15(1/32) (= –1/2)
E(–Y/2) = –32/2 = –16 oe for B1A12 In a game of chance there are 32 slots, numbered 1 to 32, and on each turn a ball lands in one of them. You may assume that the process is completely random.
You are given that $X$ is the random variable denoting the number of the slot that the ball lands in on a given turn.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable distribution to model $X$. You should state the value(s) of any parameter(s).
\item Write down $\mathrm { P } ( X = 7 )$.
Players of the game start with a score of 0 . On each turn a player may choose to play the game by selecting a number. If the ball lands in the slot with that number then 15 is added to the player's score. Otherwise, the player's score is reduced by 1 . A player's score may become negative.
A player decides to play the game, selecting the number 7 on each turn, until the ball lands in the slot numbered 7.
You are given that $Y$ is the random variable denoting the number of turns up to and including the turn in which the ball lands in the slot numbered 7.
\item Determine $\mathrm { P } ( Y \leqslant 15 )$.
\item Determine the player's expected final score.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2024 Q2 [9]}}