| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pure expectation and variance calculation |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for sums and differences of random variables. Part (a) requires routine use of formulas for means and variances, with the only slight complication being the correlation coefficient in Var(F) and Var(D). Part (b) involves standard normal probability calculations. The question is slightly easier than average because it's entirely procedural with no problem-solving or insight required—students just apply memorized formulas systematically. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Task | Mean | Standard deviation |
| \(U\) | 15 | 5 |
| \(V\) | 40 | 15 |
| \(W\) | 75 | 20 |
| \(X\) | 20 | 10 |
| Answer | Marks | Guidance |
|---|---|---|
| \(R:\) mean \(= 35\) | B1 | Both CAO |
| variance \(= 125\) | ||
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F:\) mean \(= 115\) | B1 | CAO |
| variance \(= 15^2 + 20^2 + (2 \times 15 \times 20 \times 0.25)\) | M1 | Attempt at \(a^2 + b^2 \pm (2) \times a \times b \times 0.25\) |
| \(= \mathbf{775}\) | A1 | CAO |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T:\) mean \(= 150\) | B1 | CAO |
| variance \(= 900\) | A1 | CAO |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(D:\) mean \(= 35\) | B1 | CAO |
| variance \(= 20^2 + 15^2 - (2 \times 20 \times 15 \times 0.25)\) | (M1) | Only if M1 not scored in (ii) |
| or \(= (ii) - 4 \times 15 \times 20 \times 0.25\) | ||
| \(= \mathbf{475}\) | B1 | CAO |
| 2 | ||
| 8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T < 180) = P\left(Z < \frac{180 - 150}{\sqrt{900}}\right)\) | M1 | Standardising 180 with values from (a)(iii) but must involve \(\sqrt{}\) |
| \(= P(Z < 1) = \mathbf{0.841}\) | A1 | AWRT (0.84134) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(W - V > 60) =\) | ||
| \(P\left(D > 60\right) = P\left(Z > \frac{60 - 35}{\sqrt{475}}\right)\) | M1 | Standardising 60 with values from (a)(iv) but must involve \(\sqrt{}\) |
| \(= P(Z > 1.1147) = 1 - P(Z < 1.1147)\) | M1 | Area change; can be implied by any final answer \(< 0.5\) |
| \(= 1 - (0.873 \text{ to } 0.875) = \mathbf{0.125 \text{ to } 0.127}\) | A1 | AWFW (0.12567) |
| 3 | ||
| 5 | ||
| Total | 13 |
## Part (a)(i)
$R:$ mean $= 35$ | B1 | Both CAO
variance $= 125$ | |
| | **1** |
## Part (a)(ii)
$F:$ mean $= 115$ | B1 | CAO
variance $= 15^2 + 20^2 + (2 \times 15 \times 20 \times 0.25)$ | M1 | Attempt at $a^2 + b^2 \pm (2) \times a \times b \times 0.25$
$= \mathbf{775}$ | A1 | CAO
| | **3** |
## Part (a)(iii)
$T:$ mean $= 150$ | B1 | CAO
variance $= 900$ | A1 | CAO
| | **2** |
## Part (a)(iv)
$D:$ mean $= 35$ | B1 | CAO
variance $= 20^2 + 15^2 - (2 \times 20 \times 15 \times 0.25)$ | (M1) | Only if M1 not scored in (ii)
or $= (ii) - 4 \times 15 \times 20 \times 0.25$ | |
$= \mathbf{475}$ | B1 | CAO
| | **2** |
| | **8** |
## Part (b)(i)
$P(T < 180) = P\left(Z < \frac{180 - 150}{\sqrt{900}}\right)$ | M1 | Standardising 180 with values from (a)(iii) but must involve $\sqrt{}$
$= P(Z < 1) = \mathbf{0.841}$ | A1 | AWRT (0.84134)
| | **2** |
## Part (b)(ii)
$P(W - V > 60) =$ | |
$P\left(D > 60\right) = P\left(Z > \frac{60 - 35}{\sqrt{475}}\right)$ | M1 | Standardising 60 with values from (a)(iv) but must involve $\sqrt{}$
$= P(Z > 1.1147) = 1 - P(Z < 1.1147)$ | M1 | Area change; can be implied by any final answer $< 0.5$
$= 1 - (0.873 \text{ to } 0.875) = \mathbf{0.125 \text{ to } 0.127}$ | A1 | AWFW (0.12567)
| | **3** |
| | **5** |
**Total** | | **13** |
---Ben is a fencing contractor who is often required to repair a garden fence by replacing a broken post between fence panels, as illustrated.
\includegraphics{figure_4}
The tasks involved are as follows.
$U$: detach the two fence panels from the broken post
$V$: remove the broken post
$W$: insert a new post
$X$: attach the two fence panels to the new post
The mean and the standard deviation of the time, in minutes, for each of these tasks are shown in the table.
\begin{tabular}{|c|c|c|}
\hline
Task & Mean & Standard deviation \\
\hline
$U$ & 15 & 5 \\
$V$ & 40 & 15 \\
$W$ & 75 & 20 \\
$X$ & 20 & 10 \\
\hline
\end{tabular}
The random variables $U$, $V$, $W$ and $X$ are pairwise independent, except for $V$ and $W$ for which $\rho_{VW} = 0.25$.
\begin{enumerate}[label=(\alph*)]
\item Determine values for the mean and the variance of:
\begin{enumerate}[label=(\roman*)]
\item $R = U + X$;
\item $F = V + W$;
\item $T = R + F$;
\item $D = W - V$.
\end{enumerate}
[8 marks]
\item Assuming that each of $R$, $F$, $T$ and $D$ is approximately normally distributed, determine the probability that:
\begin{enumerate}[label=(\roman*)]
\item the total time taken by Ben to repair a garden fence is less than 3 hours;
\item the time taken by Ben to insert a new post is at least 1 hour more than the time taken by him to remove the broken post.
\end{enumerate}
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2016 Q4 [13]}}