| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Session | Specimen |
| Marks | 18 |
| 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 multi-part question on discrete probability distributions requiring standard techniques: completing a table by substitution, finding a constant using ΣP=1, calculating E(X) and Var(X) using definitions, and applying basic probability counting with dice. While lengthy (7 parts), each component is routine textbook material with clear scaffolding and no novel insight required. Slightly easier than average due to the guided structure and standard methods throughout. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02a 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 | 6 |
| \(\mathrm { P } ( X = r )\) |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | r 1 2 3 4 5 6 |
| P(X = r) k 3k 5k 7k 9k 11k | B1 | |
| [1] | 1.1 | |
| 4 | (ii) | k + 3k + 5k + 7k + 9k +11k = 1 |
| Answer | Marks |
|---|---|
| 36 | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| 1.1 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (iii) | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| I1.1 | M |
| Answer | Marks | Guidance |
|---|---|---|
| r | 1 | 2 |
| P(X = r) | k | 3k |
| 4 | (iv) | DR |
| Answer | Marks |
|---|---|
| 36 (cid:169) 36 (cid:185) 1296 | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (v) | E |
| Answer | Marks |
|---|---|
| 36 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 2.1 | |
| 1.1 | May be in a sample space | |
| 4 | (vi) | Finding probability distribution for score |
| Answer | Marks |
|---|---|
| Probabilities are consecutive odd numbers out of 36 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2a | Strategy e.g. use of sample space |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (vii) | Total(cid:32) X (cid:14)X (cid:14)X |
| Answer | Marks |
|---|---|
| 1296 1296 1296 | C |
| Answer | Marks |
|---|---|
| [3] | I |
| Answer | Marks |
|---|---|
| 1.1 | FT from (iv) |
| Answer | Marks | Guidance |
|---|---|---|
| Score | 1 | 2 |
| 4 | 4 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | 1 | 1 |
| 0 | 0 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(iii) | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(iv) | 5 | 0 |
| 0 | 0 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(v) | 1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(vi) | 1 | C |
| 2 | 0 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(vii) | 3 | 0 |
Question 4:
4 | (i) | r 1 2 3 4 5 6
P(X = r) k 3k 5k 7k 9k 11k | B1
[1] | 1.1
4 | (ii) | k + 3k + 5k + 7k + 9k +11k = 1
1
36k = 1 so k =
36
1
so k = AG
36 | M1
A1
[2] | 2.4
1.1 | N
For equation
E
4 | (iii) | B1
B1
C
[2] | 1.1
I1.1 | M
For heights
For axes and labels
r | 1 | 2 | 3 | 4 | 5 | 6
P(X = r) | k | 3k | 5k | 7k | 9k | 11k
4 | (iv) | DR
1 2 5 7 9
E(X)(cid:32)1(cid:117) (cid:14) 2(cid:117) (cid:14) 3(cid:117) (cid:14)4(cid:117) (cid:14)5(cid:117)
36 36 36 36 36
11
(cid:14)6(cid:117)
36
161
(cid:32) (cid:62)(cid:32)4.47(cid:64)
36
E (cid:11) X2(cid:12) (cid:32)12(cid:117) 1 +22(cid:117) 3 (cid:14)32(cid:117) 5 (cid:14)42(cid:117) 7 (cid:14)52(cid:117) 9
36 36 36 36 36
11
(cid:14)62(cid:117)
36
791
(cid:32)
36
2
791 (cid:167)161(cid:183) 2555
Var(cid:11)X(cid:12) (cid:32) – (cid:168) (cid:184) (cid:32) (cid:11)(cid:32) 1.97(cid:12)
36 (cid:169) 36 (cid:185) 1296 | M1
A1
M1
C
M1
A1
[5] | 1.1a
1.1
1.1
I
1.2
1.1 | N
E
M
4 | (v) | E
Combinations which lead to score of 3 are
P
(1, 3), (2, 3), (3, 3), (3, 2), (3, 1)
36 possible outcomes
S
5
so Probability = AG
36 | M1
A1
[2] | 2.1
1.1 | May be in a sample space
4 | (vi) | Finding probability distribution for score
Score 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 2 3 4 5 6
3 3 3 3 4 5 6
4 4 4 4 4 5 6
5 5 5 5 5 5 6
6 6 6 6 6 6 6
Probabilities are consecutive odd numbers out of 36 | B1
B1
E1
[3] | 1.1
2.1
2.2a | Strategy e.g. use of sample space
N
Working through strategy
e.g. completing sample space
E
M
Matching probabilities for score
with those for X. Must be
convincing
4 | (vii) | Total(cid:32) X (cid:14)X (cid:14)X
1 2 3
161 161 161 483
E(X (cid:14)X (cid:14)X ) (cid:32) (cid:14) (cid:14) (cid:32) ((cid:32)13.42)
1 2 3
36 36 36 36E
2555 2555 2555
Var(X (cid:14)X (cid:14)X ) (cid:32) (cid:14) (cid:14) (cid:32) 5.91
1 2 3
1296 1296 1296 | C
B1
B1
B1
[3] | I
1.1a
1.1
1.1 | FT from (iv)
FT from (iv)
Score | 1 | 2 | 3 | 4 | 5 | 6
4 | 4 | 4 | 4 | 4 | 5 | 6
--- 4(i) ---
4(i) | 1 | 0 | 0 | 0 | 1
--- 4(ii) ---
4(ii) | 1 | 1 | M
0 | 0 | 2
--- 4(iii) ---
4(iii) | 2 | 0 | 0 | 0 | 2
--- 4(iv) ---
4(iv) | 5 | 0 | I
0 | 0 | 5
--- 4(v) ---
4(v) | 1 | 1 | 0 | 0 | 2
--- 4(vi) ---
4(vi) | 1 | C
2 | 0 | 0 | 3
--- 4(vii) ---
4(vii) | 3 | 0 | 0 | 0 | 34 The discrete random variable $X$ has probability distribution defined by
$$\mathrm { P } ( X = r ) = k ( 2 r - 1 ) \quad \text { for } r = 1,2,3,4,5,6 \text {, where } k \text { is a constant. }$$
(i) Complete the table in the Printed Answer Booklet giving the probabilities in terms of $k$.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | }
\hline
$r$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( X = r )$ & & & & & & \\
\hline
\end{tabular}
\end{center}
(ii) Show that the value of $k$ is $\frac { 1 } { 36 }$.\\
(iii) Draw a graph to illustrate the distribution.\\
(iv) In this question you must show detailed reasoning.
Find
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$.
\end{itemize}
A game consists of a player throwing two fair dice. The score is the maximum of the two values showing on the dice.\\
(v) Show that the probability of a score of 3 is $\frac { 5 } { 36 }$.\\
(vi) Show that the probability distribution for the score in the game is the same as the probability distribution of the random variable $X$.\\
(vii) The game is played three times.
Find
\begin{itemize}
\item the mean of the total of the three scores.
\item the variance of the total of the three scores.
\end{itemize}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS Q4 [18]}}