| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Standard +0.3 This is a straightforward application of normal distribution properties and linear combinations. Parts (a) and (b) are standard normal probability calculations, (c) requires forming a linear combination of normals (a core Further Stats topic but routine), and (d) applies sampling distribution of means. All steps are textbook procedures with no novel insight required, though it's slightly above average due to being Further Maths content and requiring multiple techniques. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| \cline { 2 - 3 } \multicolumn{1}{c|}{} | Mean | Standard deviation |
| Washing | 35 | 2.4 |
| Drying | 46 | 3.1 |
| Folding | 12 | 2.2 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | N(46, 3.12) P(> 50 mins) = 0.0985 |
| [1] | 1.1 | BC |
| 3 | (b) | N(35, 2.42) inv(0.99) 40.6 mins |
| [1] | 3.1b | BC |
| 3 | (c) | N(46 – 35 – 12, 3.12 + 2.42 + 2.22 ) |
| Answer | Marks |
|---|---|
| P( < 0) = 0.5880 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | For Normal and mean. Allow mean = +1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (d) | Time for 1 load ~ N(35 + 46 + 12, 2.42 + 3.12 + 2.22) |
| Answer | Marks |
|---|---|
| So P(Mean < 90) = 0.0678 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | For correct distribution soi |
Question 3:
3 | (a) | N(46, 3.12) P(> 50 mins) = 0.0985 | B1
[1] | 1.1 | BC
3 | (b) | N(35, 2.42) inv(0.99) 40.6 mins | B1
[1] | 3.1b | BC
3 | (c) | N(46 – 35 – 12, 3.12 + 2.42 + 2.22 )
N(−1, 20.21)
P( < 0) = 0.5880 | B1
B1
B1
[3] | 3.3
1.1
3.4 | For Normal and mean. Allow mean = +1
For variance (4.4955…2) awrt 20.2
BC
3 | (d) | Time for 1 load ~ N(35 + 46 + 12, 2.42 + 3.12 + 2.22)
(N(93, 20.21) )
Mean time for 5 loads ~ N(93, 20.21/5)
So P(Mean < 90) = 0.0678 | B1
M1
A1
[3] | 3.3
1.1
3.4 | For correct distribution soi
For distribution
BC3 At a launderette the process of cleaning a load of clothes consists of three stages: washing, drying and folding. The times in minutes for each process are modelled by independent Normal distributions with means and standard deviations as shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | }
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & Mean & Standard deviation \\
\hline
Washing & 35 & 2.4 \\
\hline
Drying & 46 & 3.1 \\
\hline
Folding & 12 & 2.2 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the probability that drying a randomly chosen load of clothes takes more than 50 minutes.
\item It is given that for $99 \%$ of loads of clothes the washing time is less than $k$ minutes.
Find the value of $k$.
\item Determine the probability that the drying time for a randomly chosen load of clothes is less than the total of the washing and folding times.
\item Determine the probability that the mean time for cleaning 5 randomly chosen loads of clothes is less than 90 minutes. You should assume that the time for cleaning any load is independent of the time for cleaning any other load.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2024 Q3 [8]}}