| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Expectation and variance with context application |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for linear combinations of random variables. Part (a) requires direct use of E(aX+bY) and Var(aX+bY) formulas with correlation, which are bookwork for S3. Part (b) involves summing independent normal variables and a routine normal probability calculation. The context is clear and all necessary information is provided, making this slightly easier than average for an S3 question. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(T) = 2E(W) + 2E(H) = 2(350) + 2(210) = 1120\) | B1 | |
| \(Var(T) = 4Var(W) + 4Var(H) + 2(2)(2)Cov(W,H)\) | M1 | Correct formula attempt including covariance term |
| \(Cov(W,H) = \rho_{WH}\sqrt{Var(W)}\sqrt{Var(H)} = 0.75\sqrt{5}\sqrt{4}\) | M1 | Correct use of correlation to find covariance |
| \(= 0.75 \times \sqrt{5} \times 2 = 3\sqrt{5}\) | A1 | |
| \(Var(T) = 4(5) + 4(4) + 8(3\sqrt{5}/\sqrt{5}\cdot\sqrt{5})\)... \(= 20 + 16 + 8(0.75)(2\sqrt{5}/\sqrt{1})\) | ||
| \(Var(T) = 4(5) + 4(4) + 8 \times 0.75\sqrt{20} = 36 + 8(0.75)(2\sqrt{5})\) | ||
| \(Var(T) = 20 + 16 + 8 \times \frac{3\sqrt{5}\cdot\sqrt{5}\cdot\sqrt{4}}{\sqrt{5}}\)... \(= 36 + 8(1.5\sqrt{5})\) | ||
| \(Var(T) = 20 + 16 + 8 \times 0.75 \times \sqrt{20} \approx 56 + 26.83 \approx 82.83\) | A1 | Accept exact form \(56 + 12\sqrt{20}\) or \(56+24\sqrt{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total tape \(S = T_1 + T_2 + T_3 + T_4\), drawers independent | M1 | |
| \(E(S) = 4 \times 1120 = 4480\) mm | B1 | ft their E(T) |
| \(Var(S) = 4 \times Var(T)\) | M1 | Using independence |
| \(= 4 \times 82.83... \approx 331.3\) | A1 | ft their Var(T) |
| \(P(S \leq 4500) = P\left(Z \leq \frac{4500 - 4480}{\sqrt{331.3}}\right) = P(Z \leq 1.099...)\) | M1 | Standardising, converting 4.5m to 4500mm |
| \(\approx \Phi(1.10) \approx 0.8643\) | A1 | Accept answers in range 0.863–0.866 |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(T) = 2E(W) + 2E(H) = 2(350) + 2(210) = 1120$ | B1 | |
| $Var(T) = 4Var(W) + 4Var(H) + 2(2)(2)Cov(W,H)$ | M1 | Correct formula attempt including covariance term |
| $Cov(W,H) = \rho_{WH}\sqrt{Var(W)}\sqrt{Var(H)} = 0.75\sqrt{5}\sqrt{4}$ | M1 | Correct use of correlation to find covariance |
| $= 0.75 \times \sqrt{5} \times 2 = 3\sqrt{5}$ | A1 | |
| $Var(T) = 4(5) + 4(4) + 8(3\sqrt{5}/\sqrt{5}\cdot\sqrt{5})$... $= 20 + 16 + 8(0.75)(2\sqrt{5}/\sqrt{1})$ | | |
| $Var(T) = 4(5) + 4(4) + 8 \times 0.75\sqrt{20} = 36 + 8(0.75)(2\sqrt{5})$ | | |
| $Var(T) = 20 + 16 + 8 \times \frac{3\sqrt{5}\cdot\sqrt{5}\cdot\sqrt{4}}{\sqrt{5}}$... $= 36 + 8(1.5\sqrt{5})$ | | |
| $Var(T) = 20 + 16 + 8 \times 0.75 \times \sqrt{20} \approx 56 + 26.83 \approx 82.83$ | A1 | Accept exact form $56 + 12\sqrt{20}$ or $56+24\sqrt{5}$ |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Total tape $S = T_1 + T_2 + T_3 + T_4$, drawers independent | M1 | |
| $E(S) = 4 \times 1120 = 4480$ mm | B1 | ft their E(T) |
| $Var(S) = 4 \times Var(T)$ | M1 | Using independence |
| $= 4 \times 82.83... \approx 331.3$ | A1 | ft their Var(T) |
| $P(S \leq 4500) = P\left(Z \leq \frac{4500 - 4480}{\sqrt{331.3}}\right) = P(Z \leq 1.099...)$ | M1 | Standardising, converting 4.5m to 4500mm |
| $\approx \Phi(1.10) \approx 0.8643$ | A1 | Accept answers in range 0.863–0.866 |
---5 In the manufacture of desk drawer fronts, a machine cuts sheets of veneered chipboard into rectangular pieces of width $W$ millimetres and height $H$ millimetres. The 4 edges of each of these pieces are then covered with matching veneered tape.
The distributions of $W$ and $H$ are such that
$$\mathrm { E } ( W ) = 350 \quad \operatorname { Var } ( W ) = 5 \quad \mathrm { E } ( H ) = 210 \quad \operatorname { Var } ( H ) = 4 \quad \rho _ { W H } = 0.75$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean and the variance of the length of tape, $T = 2 W + 2 H$, needed for the edges of a drawer front.
\item A desk has 4 such drawers whose sizes may be assumed to be independent.
Given that $T$ may be assumed to be normally distributed, determine the probability that the total length of tape needed for the edges of the desk's 4 drawer fronts does not exceed 4.5 metres.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-13_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2010 Q5 [10]}}