| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 18 |
| 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 normal random variables. All parts require only direct use of formulas for means and variances of sums/differences of normals, followed by routine normal probability calculations. No novel insight or complex problem-solving is needed—just careful arithmetic and correct application of well-practiced techniques from the S3 syllabus. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Vehicle type | Mean | Standard deviation |
| Cars | 60.2 | 5.2 |
| Coaches | 33.9 | 6.3 |
| Lorries | 52.4 | 4.9 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Co < 40) = P(Z < \frac{40-33.9}{6.3}) = P(Z < 0.9683) = 0.8336\) | M1, A1, A1 | For standardising. Award once, here or elsewhere. c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(L - Ca \sim N(52.4 - 60.2 = -7.8, 4.9^2 + 5.2^2 = 51.05)\) | M1, B1, B1 | Allow \(Ca - L\) provided subsequent work is consistent. Mean. Variance. Accept \(sd = \sqrt{51.05} = 7.1449...\) |
| \(P(\text{this} > 0) = P(Z > \frac{0-(-7.8)}{\sqrt{51.05}}) = P(Z > 1.0917) = 1 - 0.8625 = 0.1375\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(Ca_1 + ... \sim N(60.2 + 60.2 + 60.2 + 60.2 = 240.8, 5.2^2 + 5.2^2 + 5.2^2 + 5.2^2 = 108.16)\) | M1, B1, B1 | Mean. Variance. Accept \(sd = \sqrt{108.16} = 10.4\). |
| \(P(\text{this} > 225) = P(Z > \frac{225-240.8}{\sqrt{108.16}}) = P(Z > -1.519) = 0.9356\) | A1 | c.a.o. |
| Must assume that the weeks are independent of each other. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(R \sim N(0.05 \times 60.2 + 0.1 \times 33.9 + 0.2 \times 52.4 = 16.88, 0.05^2 \times 5.2^2 + 0.1^2 \times 6.3^2 + 0.2^2 \times 4.9^2 = 1.4249)\) | M1, A1, M1, M1, A1 | Mean. For \(0.05^2\) etc. For \(\times 5.2^2\) etc. Accept \(sd = \sqrt{1.4249} = 1.1937\). |
| \(P(R > 20) = P(Z > \frac{20-16.88}{\sqrt{1.4249}}) = P(Z > 2.613) = 1 - 0.9955 = 0.0045\) | A1 | c.a.o. |
**Part (i):**
$P(Co < 40) = P(Z < \frac{40-33.9}{6.3}) = P(Z < 0.9683) = 0.8336$ | M1, A1, A1 | For standardising. Award once, here or elsewhere. c.a.o. | 3 marks
**Part (ii):**
Want $P(L > Ca)$ i.e. $P(L - Ca > 0)$
$L - Ca \sim N(52.4 - 60.2 = -7.8, 4.9^2 + 5.2^2 = 51.05)$ | M1, B1, B1 | Allow $Ca - L$ provided subsequent work is consistent. Mean. Variance. Accept $sd = \sqrt{51.05} = 7.1449...$
$P(\text{this} > 0) = P(Z > \frac{0-(-7.8)}{\sqrt{51.05}}) = P(Z > 1.0917) = 1 - 0.8625 = 0.1375$ | A1 | c.a.o. | 4 marks
**Part (iii):**
Want $P(Ca_1 + Ca_2 + Ca_3 + Ca_4 > 225)$
$Ca_1 + ... \sim N(60.2 + 60.2 + 60.2 + 60.2 = 240.8, 5.2^2 + 5.2^2 + 5.2^2 + 5.2^2 = 108.16)$ | M1, B1, B1 | Mean. Variance. Accept $sd = \sqrt{108.16} = 10.4$.
$P(\text{this} > 225) = P(Z > \frac{225-240.8}{\sqrt{108.16}}) = P(Z > -1.519) = 0.9356$ | A1 | c.a.o.
Must assume that the weeks are independent of each other. | B1 | | 5 marks
**Part (iv):**
$R \sim N(0.05 \times 60.2 + 0.1 \times 33.9 + 0.2 \times 52.4 = 16.88, 0.05^2 \times 5.2^2 + 0.1^2 \times 6.3^2 + 0.2^2 \times 4.9^2 = 1.4249)$ | M1, A1, M1, M1, A1 | Mean. For $0.05^2$ etc. For $\times 5.2^2$ etc. Accept $sd = \sqrt{1.4249} = 1.1937$.
$P(R > 20) = P(Z > \frac{20-16.88}{\sqrt{1.4249}}) = P(Z > 2.613) = 1 - 0.9955 = 0.0045$ | A1 | c.a.o. | 6 marks2 The operator of a section of motorway toll road records its weekly takings according to the types of vehicles using the motorway. For purposes of charging, there are three types of vehicle: cars, coaches, lorries. The weekly takings (in thousands of pounds) for each type are assumed to be Normally distributed. These distributions are independent of each other and are summarised in the table.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
Vehicle type & Mean & Standard deviation \\
\hline
Cars & 60.2 & 5.2 \\
\hline
Coaches & 33.9 & 6.3 \\
\hline
Lorries & 52.4 & 4.9 \\
\hline
\end{tabular}
\end{center}
(i) Find the probability that the weekly takings for coaches are less than $\pounds 40000$.\\
(ii) Find the probability that the weekly takings for lorries exceed the weekly takings for cars.\\
(iii) Find the probability that over a 4 -week period the total takings for cars exceed $\pounds 225000$. What assumption must be made about the four weeks?\\
(iv) Each week the operator allocates part of the takings for repairs. This is determined for each type of vehicle according to estimates of the long-term damage caused. It is calculated as follows: $5 \%$ of takings for cars, $10 \%$ for coaches and $20 \%$ for lorries. Find the probability that in any given week the total amount allocated for repairs will exceed $\pounds 20000$.
\hfill \mbox{\textit{OCR MEI S3 2007 Q2 [18]}}