| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Multiple independent trials or dice |
| Difficulty | Easy -1.2 This question tests basic discrete uniform distribution properties with straightforward formula application. Part (a) is simple counting (8/21), part (b) uses standard formulas for mean and variance that are typically given in formula books. The spreadsheet context in the stem doesn't add difficulty to these particular parts, which require only routine recall and calculation. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02e Discrete uniform distribution |
| \includegraphics[max width=\textwidth, alt={}]{77eabbd6-a058-457f-9601-d66f3c2db005-07_38_45_880_279} | A | B | C | D | E | F | G | H | I | J | K | L |
| 1 | \(X _ { 1 }\) | \(X _ { 2 }\) | \(X _ { 3 }\) | \(X _ { 4 }\) | \(X _ { 5 }\) | \(X _ { 6 }\) | \(X _ { 7 }\) | \(X _ { 8 }\) | \(X _ { 9 }\) | \(X _ { 10 }\) | \(Y\) | |
| 2 | 1 | 6 | 2 | 1 | 18 | 6 | 4 | 9 | 11 | 11 | 6.9 | |
| 3 | 13 | 14 | 12 | 2 | 4 | 11 | 16 | 0 | 16 | 0 | 8.8 | |
| 4 | 4 | 17 | 1 | 16 | 4 | 10 | 12 | 2 | 18 | 13 | 9.7 | |
| 5 | 2 | 8 | 12 | 1 | 4 | 16 | 12 | 2 | 15 | 8 | 8.0 | |
| 6 | 7 | 15 | 16 | 0 | 4 | 7 | 1 | 13 | 0 | 20 | 8.3 | |
| 7 | 15 | 13 | 10 | 1 | 12 | 0 | 20 | 15 | 16 | 6 | 10.8 | |
| 8 | 14 | 13 | 17 | 12 | 2 | 18 | 16 | 18 | 9 | 4 | 12.3 | |
| 9 | 20 | 2 | 12 | 3 | 17 | 3 | 0 | 18 | 15 | 13 | 10.3 | |
| 10 | 2 | 12 | 5 | 12 | 2 | 6 | 0 | 9 | 10 | 15 | 7.3 | |
| 11 | 5 | 11 | 13 | 10 | 9 | 17 | 10 | 4 | 20 | 15 | 11.4 | |
| 12 | 14 | 9 | 9 | 7 | 6 | 20 | 2 | 2 | 11 | 16 | 9.6 | |
| 13 | 15 | 19 | 18 | 19 | 7 | 6 | 6 | 20 | 3 | 8 | 12.1 | |
| 14 | 5 | 10 | 6 | 4 | 1 | 19 | 15 | 8 | 17 | 18 | 10.3 | |
| 15 | 0 | 3 | 15 | 15 | 11 | 12 | 0 | 3 | 9 | 16 | 8.4 | |
| 16 | 1 | 12 | 1 | 15 | 0 | 4 | 11 | 11 | 9 | 2 | 6.6 | |
| 17 | 12 | 5 | 0 | 8 | 3 | 8 | 12 | 19 | 13 | 12 | 9.2 | |
| 18 | 9 | 5 | 1 | 13 | 5 | 4 | 18 | 1 | 1 | 19 | 7.6 | |
| 19 | 16 | 2 | 20 | 20 | 12 | 17 | 2 | 7 | 8 | 20 | 12.4 | |
| 20 | 18 | 17 | 3 | 2 | 8 | 18 | 7 | 0 | 11 | 6 | 9.0 | |
| 21 | 15 | 10 | 7 | 20 | 4 | 0 | 5 | 6 | 11 | 14 | 9.2 | |
| 22 | 3 | 9 | 10 | 14 | 2 | 1 | 8 | 6 | 0 | 7 | 6.0 | |
| 23 | 11 | 10 | 11 | 10 | 19 | 11 | 3 | 7 | 10 | 0 | 9.2 | |
| 24 | 12 | 14 | 6 | 6 | 5 | 20 | 11 | 18 | 10 | 14 | 11.6 | |
| 25 | 1 | 11 | 5 | 14 | 11 | 10 | 1 | 1 | 2 | 0 | 5.6 | |
| 26 | 0 | 14 | 7 | 11 | 18 | 5 | 10 | 20 | 11 | 9 | 10.5 | |
| 27 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (a) | P(X ≤7)= 8 |
| 21 | B1 | |
| [1] | 1.1 | Allow 0.381 |
| 9 | (b) | E(X) = 10 |
| Answer | Marks |
|---|---|
| 3 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Allow M1 for Var(X)= 1 (202 −1) |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (c) | P(Y ≤7)= 4 |
| 25 | B1 | |
| [1] | 1.1 | |
| 9 | (d) | Var(X) |
| Answer | Marks |
|---|---|
| mean than values of X. (E(X) and E(Y) are both 10) | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.4 | Allow Var(Y) < Var(X) If they get Var(Y) = Var(X) /100 or |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (e) | Var(W)=11 |
| Answer | Marks |
|---|---|
| 0.00348 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Allow their |
Question 9:
9 | (a) | P(X ≤7)= 8
21 | B1
[1] | 1.1 | Allow 0.381
9 | (b) | E(X) = 10
Var(X)= 1 (212−1)
12
=110
3 | B1
M1
A1
[3] | 1.1
1.2
1.1 | Allow M1 for Var(X)= 1 (202 −1)
12
Or 36.7 or better
9 | (c) | P(Y ≤7)= 4
25 | B1
[1] | 1.1
9 | (d) | Var(X)
Because Var(Y)=
10
and so values of Y are more likely to be closer to the
mean than values of X. (E(X) and E(Y) are both 10) | B1
B1
[2] | 2.2a
2.4 | Allow Var(Y) < Var(X) If they get Var(Y) = Var(X) /100 or
other error then B0 even if then say Var(Y) < Var(X)
oe
9 | (e) | Var(W)=11
9
( 11)
By CLT distribution is approx N 10,
9
P(W ≤ 7) with CC is P ( Z ≤7 1 )
60
0.00348 | B1
M1
B1
A1
[4] | 1.1
2.2a
3.4
1.1 | Allow their
Var(𝑋𝑋)
Using their mean from (b) and their var(W)
30
For CC (Continuity correction)
Allow 3 marks for 0.00333 (0.0033278…) (No CC) Allow 0.0035
Allow equivalent method eg using total rather than mean9 The random variable $X$ has a discrete uniform distribution over the values $\{ 0,1,2 , \ldots , 20 \}$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X \leqslant 7 )$.
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\end{itemize}
The spreadsheet shows a simulation of the distribution of $X$. Each of the 25 rows of the spreadsheet below the heading row shows a simulation of 10 independent values of $X$ together with the value of the mean of the 10 values, denoted by $Y$.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
\includegraphics[max width=\textwidth, alt={}]{77eabbd6-a058-457f-9601-d66f3c2db005-07_38_45_880_279}
& A & B & C & D & E & F & G & H & I & J & K & L \\
\hline
1 & $X _ { 1 }$ & $X _ { 2 }$ & $X _ { 3 }$ & $X _ { 4 }$ & $X _ { 5 }$ & $X _ { 6 }$ & $X _ { 7 }$ & $X _ { 8 }$ & $X _ { 9 }$ & $X _ { 10 }$ & $Y$ & \\
\hline
2 & 1 & 6 & 2 & 1 & 18 & 6 & 4 & 9 & 11 & 11 & 6.9 & \\
\hline
3 & 13 & 14 & 12 & 2 & 4 & 11 & 16 & 0 & 16 & 0 & 8.8 & \\
\hline
4 & 4 & 17 & 1 & 16 & 4 & 10 & 12 & 2 & 18 & 13 & 9.7 & \\
\hline
5 & 2 & 8 & 12 & 1 & 4 & 16 & 12 & 2 & 15 & 8 & 8.0 & \\
\hline
6 & 7 & 15 & 16 & 0 & 4 & 7 & 1 & 13 & 0 & 20 & 8.3 & \\
\hline
7 & 15 & 13 & 10 & 1 & 12 & 0 & 20 & 15 & 16 & 6 & 10.8 & \\
\hline
8 & 14 & 13 & 17 & 12 & 2 & 18 & 16 & 18 & 9 & 4 & 12.3 & \\
\hline
9 & 20 & 2 & 12 & 3 & 17 & 3 & 0 & 18 & 15 & 13 & 10.3 & \\
\hline
10 & 2 & 12 & 5 & 12 & 2 & 6 & 0 & 9 & 10 & 15 & 7.3 & \\
\hline
11 & 5 & 11 & 13 & 10 & 9 & 17 & 10 & 4 & 20 & 15 & 11.4 & \\
\hline
12 & 14 & 9 & 9 & 7 & 6 & 20 & 2 & 2 & 11 & 16 & 9.6 & \\
\hline
13 & 15 & 19 & 18 & 19 & 7 & 6 & 6 & 20 & 3 & 8 & 12.1 & \\
\hline
14 & 5 & 10 & 6 & 4 & 1 & 19 & 15 & 8 & 17 & 18 & 10.3 & \\
\hline
15 & 0 & 3 & 15 & 15 & 11 & 12 & 0 & 3 & 9 & 16 & 8.4 & \\
\hline
16 & 1 & 12 & 1 & 15 & 0 & 4 & 11 & 11 & 9 & 2 & 6.6 & \\
\hline
17 & 12 & 5 & 0 & 8 & 3 & 8 & 12 & 19 & 13 & 12 & 9.2 & \\
\hline
18 & 9 & 5 & 1 & 13 & 5 & 4 & 18 & 1 & 1 & 19 & 7.6 & \\
\hline
19 & 16 & 2 & 20 & 20 & 12 & 17 & 2 & 7 & 8 & 20 & 12.4 & \\
\hline
20 & 18 & 17 & 3 & 2 & 8 & 18 & 7 & 0 & 11 & 6 & 9.0 & \\
\hline
21 & 15 & 10 & 7 & 20 & 4 & 0 & 5 & 6 & 11 & 14 & 9.2 & \\
\hline
22 & 3 & 9 & 10 & 14 & 2 & 1 & 8 & 6 & 0 & 7 & 6.0 & \\
\hline
23 & 11 & 10 & 11 & 10 & 19 & 11 & 3 & 7 & 10 & 0 & 9.2 & \\
\hline
24 & 12 & 14 & 6 & 6 & 5 & 20 & 11 & 18 & 10 & 14 & 11.6 & \\
\hline
25 & 1 & 11 & 5 & 14 & 11 & 10 & 1 & 1 & 2 & 0 & 5.6 & \\
\hline
26 & 0 & 14 & 7 & 11 & 18 & 5 & 10 & 20 & 11 & 9 & 10.5 & \\
\hline
27 & & & & & & & & & & & & \\
\hline
\end{tabular}
\end{center}
\item Use the spreadsheet to estimate $\mathrm { P } ( Y \leqslant 7 )$.
\item Explain why the true value of $\mathrm { P } ( Y \leqslant 7 )$ is less than $\mathrm { P } ( X \leqslant 7 )$, relating your answer to $\operatorname { Var } ( X )$ and $\operatorname { Var } ( Y )$.
\item The random variable $W$ is the mean of 30 independent values of $X$.
Determine an estimate of $\mathrm { P } ( W \leqslant 7 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2022 Q9 [11]}}