| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparing two journey times |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for linear combinations of random variables (means, variances with correlation, independence). Part (a) uses basic formulas including Var(X+Y) with correlation; part (b) applies independence; part (c) requires routine normal probability calculations. All steps are mechanical with no novel insight required, making it slightly easier than average for S3 level. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a)(i) | ||
| \(E(F) = 128 + 112 = 240\) | B1 | CAO |
| Part (a)(ii) | ||
| \(\text{Cov}(X, Y) = -0.4 \times \sqrt{50 \times 50} = -20\) | M1 | Used; or equivalent |
| \(\text{Var}(F) = 50 + 50 + (2 \times -20) = 60\) | M1; A1 | \(\text{V}(X) + \text{V}(Y) + 2\text{Cov}(X,Y)\) used CAO; AG; 4 marks |
| Part (b)(i) | ||
| \(E(T) = 240 + 75 = 315\) | B1↑ | ft on (a)(i) |
| \(\text{Var}(T) = 60 + 36 = 96\) | B1 | CAO; 2 marks |
| Part (b)(ii) | ||
| \(E(M) = 240 - (3 \times 75) = 15\) | B1↑ | ft on (a)(i) |
| \(\text{Var}(M) = 60 + \{(-3^2) \times 36\} = 60 + 324 = 384\) | M1; A1 | \(\text{V}(F) + 3^2\text{V}(S)\) used CAO; 3 marks |
| Part (c)(i) | ||
| \(P(T > 300) = P\left(Z > \frac{300 - 315}{\sqrt{96}}\right)\) | M1 | Standardising 300 or 300.5 using (b)(i) |
| \(= P(Z > -1.53) = P(Z < 1.53)\) | m1 | Area change |
| \(= 0.936 \text{ to } 0.938\) | A1 | AWFW; 3 marks |
| Part (c)(ii) | ||
| \(P\left(S > \frac{X+Y}{3}\right) = \) | M1 | Used; or equivalent |
| \(P(3S > X + Y) = P(3S - F) = \) | M1 | Attempt to change to \(M\) |
| \(P(F - 3S < 0) = P(M < 0)\) | A1 | Or equivalent |
| \(= P\left(Z < \frac{0 - 15}{\sqrt{384}}\right)\) | M1 | Standardising 0 using (b)(ii) |
| \(= P(Z < -0.765) = 1 - P(Z < 0.765)\) | m1 | Area change |
| \(= 0.22(0) \text{ to } 0.225\) | A1 | AWFW; 6 marks |
| **Part (a)(i)** |
|---|
| $E(F) = 128 + 112 = 240$ | B1 | CAO |
| **Part (a)(ii)** |
|---|
| $\text{Cov}(X, Y) = -0.4 \times \sqrt{50 \times 50} = -20$ | M1 | Used; or equivalent |
| $\text{Var}(F) = 50 + 50 + (2 \times -20) = 60$ | M1; A1 | $\text{V}(X) + \text{V}(Y) + 2\text{Cov}(X,Y)$ used CAO; AG; 4 marks |
| **Part (b)(i)** |
|---|
| $E(T) = 240 + 75 = 315$ | B1↑ | ft on (a)(i) |
| $\text{Var}(T) = 60 + 36 = 96$ | B1 | CAO; 2 marks |
| **Part (b)(ii)** |
|---|
| $E(M) = 240 - (3 \times 75) = 15$ | B1↑ | ft on (a)(i) |
| $\text{Var}(M) = 60 + \{(-3^2) \times 36\} = 60 + 324 = 384$ | M1; A1 | $\text{V}(F) + 3^2\text{V}(S)$ used CAO; 3 marks |
| **Part (c)(i)** |
|---|
| $P(T > 300) = P\left(Z > \frac{300 - 315}{\sqrt{96}}\right)$ | M1 | Standardising 300 or 300.5 using (b)(i) |
| $= P(Z > -1.53) = P(Z < 1.53)$ | m1 | Area change |
| $= 0.936 \text{ to } 0.938$ | A1 | AWFW; 3 marks |
| **Part (c)(ii)** |
|---|
| $P\left(S > \frac{X+Y}{3}\right) = $ | M1 | Used; or equivalent |
| $P(3S > X + Y) = P(3S - F) = $ | M1 | Attempt to change to $M$ |
| $P(F - 3S < 0) = P(M < 0)$ | A1 | Or equivalent |
| $= P\left(Z < \frac{0 - 15}{\sqrt{384}}\right)$ | M1 | Standardising 0 using (b)(ii) |
| $= P(Z < -0.765) = 1 - P(Z < 0.765)$ | m1 | Area change |
| $= 0.22(0) \text{ to } 0.225$ | A1 | AWFW; 6 marks |
**Total: 18 marks**
---6 An aircraft, based at airport A, flies regularly to and from airport B.\\
The aircraft's flying time, $X$ minutes, from A to B has a mean of 128 and a variance of 50 .\\
The aircraft's flying time, $Y$ minutes, on the return flight from B to A is such that
$$\mathrm { E } ( Y ) = 112 , \quad \operatorname { Var } ( Y ) = 50 \quad \text { and } \quad \rho _ { X Y } = - 0.4$$
\begin{enumerate}[label=(\alph*)]
\item Given that $F = X + Y$ :
\begin{enumerate}[label=(\roman*)]
\item find the mean of $F$;
\item show that the variance of $F$ is 60 .
\end{enumerate}\item At airport B , the stopover time, $S$ minutes, is independent of $F$ and has a mean of 75 and a variance of 36 .
Find values for the mean and the variance of:
\begin{enumerate}[label=(\roman*)]
\item $T = F + S$;
\item $M = F - 3 S$.
\end{enumerate}\item Hence, assuming that $T$ and $M$ are normally distributed, determine the probability that, on a particular round trip of the aircraft from A to B and back to A :
\begin{enumerate}[label=(\roman*)]
\item the time from leaving A to returning to A exceeds 300 minutes;
\item the stopover time is greater than one third of the total flying time.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2008 Q6 [18]}}