| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum versus sum comparison |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for sums of independent normal random variables. Parts (a), (c), and (d) require only direct application of formulas for mean and variance of sums, while part (b) tests understanding that pieces from the same sheet are not independent (variance remains 0.03²). All calculations are routine once the correct distributions are identified. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | 10 sheets thickness ⁓ N(10 × 3.125, 10 × 0.032) |
| Answer | Marks |
|---|---|
| P(thickness < 31) = 0.0042 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | For Normal and mean |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | Thickness ~ N(10 × 3.125, 102 × 0.032) |
| Answer | Marks |
|---|---|
| P(thickness < 31) = 0.2023 | B1 |
| B1 | 3.3 |
| 1.1 | For correct distribution |
| Answer | Marks | Guidance |
|---|---|---|
| P(10 sheets < 31) = P(1 sheet <3.1) | B1 | |
| P(thickness < 31) = 0.2023 | B1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (c) | One of each ⁓ N(3.125+3.117+3.109, 3 × 0.032) |
| Answer | Marks |
|---|---|
| P(Total ≥ 9.4) = 0.1728 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | For method |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (d) | Distribution of difference of 10 sheets of each has |
| Answer | Marks |
|---|---|
| P(difference < 0) = 0.2755 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | oe |
Question 4:
4 | (a) | 10 sheets thickness ⁓ N(10 × 3.125, 10 × 0.032)
i.e. N(31.25, 0.009)
P(thickness < 31) = 0.0042 | M1
A1
B1
[3] | 3.3
1.1
3.4 | For Normal and mean
For correct variance
BC (Exact answer 0.00420…)
4 | (b) | Thickness ~ N(10 × 3.125, 102 × 0.032)
i.e. N(31.25, 0.09)
P(thickness < 31) = 0.2023 | B1
B1 | 3.3
1.1 | For correct distribution
BC (Exact answer 0.202328……)
Alternative method
P(10 sheets < 31) = P(1 sheet <3.1) | B1
P(thickness < 31) = 0.2023 | B1 | BC
[2]
4 | (c) | One of each ⁓ N(3.125+3.117+3.109, 3 × 0.032)
i.e. N(9.351, 0.0027)
P(Total ≥ 9.4) = 0.1728 | M1
A1
B1
[3] | 3.3
1.1
1.1 | For method
BC (Exact answer 0.172839…)
4 | (d) | Distribution of difference of 10 sheets of each has
mean = 10 × 3.125 − 10 × 3.117
variance = 10 × 0.032 + 10 × 0.032
[so distribution is N(±0.08, 0.018)]
P(difference < 0) = 0.2755 | M1
M1
A1
[3] | 3.3
1.1
3.4 | oe
Method for mean
Method for variance
BC Allow 0.275 or 0.276 (exact answer 0.27549…)4 A machine manufactures batches of 100 titanium sheets. The thickness of every sheet in a batch is Normally distributed with mean $\mu \mathrm { mm }$ and standard deviation 0.03 mm . You should assume that each sheet is of uniform thickness and that the thicknesses of different sheets are independent of each other.
The values of $\mu$ for three different batches, A, B and C, are 3.125, 3.117 and 3.109 respectively.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that the total thickness of 10 sheets from Batch A is less than 31.0 mm .
\item Determine the probability that, if a single sheet from Batch A is cut into pieces and 10 of the pieces are stacked together, the total thickness of the stack is less than 31.0 mm .
\item Determine the probability that, if one sheet from each of Batches A, B and C are stacked together, the total thickness of the stack is at least 9.4 mm .
\item Determine the probability that the total thickness of 10 sheets from Batch A is less than the total thickness of 10 sheets from Batch B.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2023 Q4 [11]}}