| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Sum of independent uniforms |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for uniform distributions and sums of independent random variables. Parts (a) and (b) require only direct recall of formulas for expectation and variance of uniform distributions and their sums. Part (c) is basic data interpretation from a table. The question appears incomplete but likely asks for a normal approximation using CLT, which is also routine. No novel problem-solving or complex multi-step reasoning required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
\multirow[b]{3}{*}{
| A | B | C | ||
| X | Y | Z | |||
| 1.17 | 3.83 | 5.01 | |||
| 3 | 2.01 | 0.81 | 2.82 | ||
| 4 | 1.27 | 1.52 | 2.78 | ||
| 5 | 1.41 | 3.94 | 5.35 | ||
| 6 | 4.11 | 2.94 | 7.05 | ||
| 7 | 1.76 | 0.96 | 2.72 | ||
| 8 | 3.29 | 0.98 | 4.27 | ||
| 9 | 0.77 | 0.22 | 0.99 | ||
| 10 | 0.99 | 1.44 | 2.43 | ||
| 11 | 4.79 | 2.43 | 7.22 | ||
| 12 | 3.82 | 3.93 | 7.75 | ||
| 13 | 5.25 | 2.74 | 7.99 | ||
| 14 | 2.64 | 0.48 | 3.12 | ||
| 15 | 1.54 | 2.18 | 3.72 | ||
| 16 | 2.71 | 1.66 | 4.36 | ||
| 17 | 0.04 | 3.24 | 3.28 | ||
| 18 | 5.95 | 3.12 | 9.07 | ||
| 19 | 5.22 | 1.21 | 6.42 | ||
| 20 | 4.16 | 0.11 | 4.27 | ||
| 21 | 1.02 | 0.99 | 2.01 | ||
| 22 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (a) | E ( X ) = 3 o r E ( Y ) = 2 |
| E(Z) = 5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.3 | |
| 1.1 | Can be implied by E(Z) = 5 | |
| 10 | (b) | 16 4 36 |
| Answer | Marks |
|---|---|
| (so Ben is wrong) | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2a | For either. Must be considering variances |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (c) | Estimate of P(Z > 6) = 0.3 |
| [1] | 1.1 | |
| 10 | (d) | 52 |
| Answer | Marks |
|---|---|
| P(Total > 210) = 0.224 (0.22376…) | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | For mean FT their E(Z) |
total
Question 10:
10 | (a) | E ( X ) = 3 o r E ( Y ) = 2
E(Z) = 5 | B1
B1
[2] | 3.3
1.1 | Can be implied by E(Z) = 5
10 | (b) | 16 4 36
Var(X) = = Var(Y) = = 3
12 3 12
16 36 52 13
Var(Z) = + = =
12 12 12 3
100 25
Cts uniform on [0, 10] has var = which is very different
12 3
(so Ben is wrong) | M1
A1
B1
[3] | 1.1
1.1
2.2a | For either. Must be considering variances
13 25
Do not allow ≠ since question says well modelled.
3 3
Allow very different, not very close oe
10 | (c) | Estimate of P(Z > 6) = 0.3 | B1
[1] | 1.1
10 | (d) | 52
Distribution is N(40×5,40× )
12
520
N(200, ) or N(200,173.33)
3
P(Total > 210) = 0.224 (0.22376…) | M1
A1
B1
[3] | 3.3
1.1
1.1 | For mean FT their E(Z)
For variance FT their Var(Z) Condone without N( , )
BC Allow equivalent method eg using mean rather than
total
B0 if continuity correction attempted10 Ben takes an underground train to work and back home each day. The waiting time is defined as the time from when he reaches the station platform until he boards the train.
On his way to work the waiting time is $X$ minutes, where $X$ is modelled by a continuous uniform distribution on $[ 0,6 ]$.
On his way back from work, the waiting time is $Y$ minutes, where $Y$ is modelled by a continuous uniform distribution on [0,4].
Ben's total waiting time for both journeys is $Z$ minutes, where $Z = X + Y$. You should assume that $X$ and $Y$ are independent.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( \mathrm { Z } )$.
\item Ben thinks that $Z$ will be well modelled by a continuous uniform distribution on $[ 0,10 ]$.
By considering variances, show that he is not correct.
\item Ben's friend Jamila constructs the spreadsheet below, which shows a simulation of 20 values of $X , Y$ and $Z$. All of the values have been rounded to 2 decimal places.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\multirow[b]{3}{*}{\begin{tabular}{l}
1 \\
2 \\
\end{tabular}} & A & B & C \\
\hline
& X & Y & Z \\
\hline
& 1.17 & 3.83 & 5.01 \\
\hline
3 & 2.01 & 0.81 & 2.82 \\
\hline
4 & 1.27 & 1.52 & 2.78 \\
\hline
5 & 1.41 & 3.94 & 5.35 \\
\hline
6 & 4.11 & 2.94 & 7.05 \\
\hline
7 & 1.76 & 0.96 & 2.72 \\
\hline
8 & 3.29 & 0.98 & 4.27 \\
\hline
9 & 0.77 & 0.22 & 0.99 \\
\hline
10 & 0.99 & 1.44 & 2.43 \\
\hline
11 & 4.79 & 2.43 & 7.22 \\
\hline
12 & 3.82 & 3.93 & 7.75 \\
\hline
13 & 5.25 & 2.74 & 7.99 \\
\hline
14 & 2.64 & 0.48 & 3.12 \\
\hline
15 & 1.54 & 2.18 & 3.72 \\
\hline
16 & 2.71 & 1.66 & 4.36 \\
\hline
17 & 0.04 & 3.24 & 3.28 \\
\hline
18 & 5.95 & 3.12 & 9.07 \\
\hline
19 & 5.22 & 1.21 & 6.42 \\
\hline
20 & 4.16 & 0.11 & 4.27 \\
\hline
21 & 1.02 & 0.99 & 2.01 \\
\hline
22 & & & \\
\hline
\end{tabular}
\end{center}
Write down an estimate of $\mathrm { P } ( Z > 6 )$.
\item Use a Normal approximation to determine the probability that Ben's total waiting time when travelling to and from work on 40 days is more than 210 minutes.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2024 Q10 [9]}}