| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Mixed sum threshold probability |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for sums and differences of independent normal random variables. Part (a) requires finding the distribution of a sum of normals and a single probability calculation. Part (b) involves finding P(3A > 2B) by considering the distribution of 3A - 2B. Part (c) is a conceptual recall question about independence. All parts use direct textbook methods with no novel problem-solving required, making this slightly easier than average for Further Maths statistics. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| \cline { 2 - 3 } \multicolumn{1}{c|}{} | Time | |||
| Type | Mean |
| ||
| A | 23 | 2.8 | ||
| B | 35 | 3.6 | ||
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | Five A ⁓ N(5 × 23, 5 × 2.82) |
| Answer | Marks |
|---|---|
| P(≥ 120) = 0.212 (0.212262…) | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | For N and mean |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | Three A – two B⁓ N(3×23 – 2×35, 3×2.82 + 2×3.62) |
| Answer | Marks |
|---|---|
| P(3 A last longer) = 0.443 (0.443453 …) | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | For N and mean Allow mean = +1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (c) | Because adding variances only gives the correct |
| combined variance if the variables are independent. | E1 | |
| [1] | 2.3 | Ignore comments about Expectation |
Question 2:
2 | (a) | Five A ⁓ N(5 × 23, 5 × 2.82)
N(115, 39.2)
P(≥ 120) = 0.212 (0.212262…) | B1
M1
A1
[3] | 3.3
1.1
3.4 | For N and mean
For variance
BC
2 | (b) | Three A – two B⁓ N(3×23 – 2×35, 3×2.82 + 2×3.62)
N(–1, 49.44)
P(3 A last longer) = 0.443 (0.443453 …) | B1
M1
A1
[3] | 3.3
1.1
3.4 | For N and mean Allow mean = +1
For variance
BC
2 | (c) | Because adding variances only gives the correct
combined variance if the variables are independent. | E1
[1] | 2.3 | Ignore comments about Expectation2 A manufacturer is testing how long coloured LED lights will last before the battery runs out, using two different battery types. The times in hours before the battery runs out are modelled by independent Normal distributions with means and standard deviations as shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | }
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & \multicolumn{2}{c|}{Time} \\
\hline
Type & Mean & \begin{tabular}{ c }
Standard \\
deviation \\
\end{tabular} \\
\hline
A & 23 & 2.8 \\
\hline
B & 35 & 3.6 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item In a particular test, a battery of type A is used and the time taken for it to run out is recorded. This process is repeated until a total of 5 randomly selected batteries have been used.
Determine the probability that the total time the 5 batteries last is at least 120 hours.
\item In a similar test, 3 randomly selected batteries of type A are used, one after the other. Then 2 randomly selected batteries of type B are used, one after the other.
Determine the probability that the 3 type A batteries last longer in total than the 2 type B batteries.
\item Explain why it is necessary that the Normal distributions are independent in order to be able to find the probability in part (b).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2022 Q2 [7]}}