| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear transformation of normal |
| Difficulty | Moderate -0.8 This is a straightforward S1 normal distribution question requiring standard z-score calculations and a simple linear transformation (multiplying by price). Part (c) asks for contextual reasoning about model validity, which is routine for S1. All techniques are standard textbook exercises with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Volume, \(X \sim N(32, 10^2)\) | ||
| \(P(X < 40) = P\left(Z < \frac{40-32}{10}\right)\) | M1 | Standardising 40 with 32 and 10; allow \((32-40)\) |
| \(= P(Z < 0.8)\) | A1 | CAO; ignore inequality and sign. May be implied by a correct answer |
| \(= 0.788\) | A1 | AWRT \((0.78814)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X > 25) = P(Z > -0.7)\) | ||
| \(= P(Z < +0.7)\) | M1 | Area change. May be implied by a correct answer or an answer \(> 0.5\) |
| \(= 0.758\) | A1 | AWRT \((0.75804)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(25 < X < 40) = \text{(i)} - (1 - \text{(ii)})\) | M1 | OE; allow new start ignoring (i) & (ii). Allow even if incorrect standardising providing \(0 <\) answer \(< 1\). May be implied by a correct answer |
| \(= 0.78814 - (1 - 0.75804) = 0.546\) | A1 | AWRT \((0.54618)\). Note: If (ii) is \(0.242\), then \((0.788 - 0.242) = 0.546 \Rightarrow\) M0 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B > \pounds 65) = P\left(Z > \frac{48.5-32}{10}\right)\) or \(P\left(Z > \frac{65-42.88}{13.4}\right)\) | M1 | Attempt to change from \(B\) to \(X\) using \((48\) to \(49)\), \(32\) and \(10\), or attempt to work with distribution of \(B\) using \(65\), \((42.8\) to \(42.9)\) and \(13.4\) |
| \(= P(Z > 1.65) = 1 - P(Z < 1.65)\) | m1 | Area change. May be implied by a correct answer or an answer \(< 0.5\) |
| \(= 1 - 0.95053 = 0.049\) to \(0.05(0)\) | A1 | AWFW \((0.04947)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Other fuels; Other vehicles with an example (not other cars); Other types of customer; Minimum purchase (policy); Purchases in integer/fixed £s; Customers filling fuel cans | B2,1 | Size of car/engine/fuel tank \(\Rightarrow\) B0; Price of fuel \(\Rightarrow\) B0; Customer paying capacity \(\Rightarrow\) B0. Must be two clearly different valid reasons for award of B2. Drivers and vehicles related \(\Rightarrow\) B1, e.g. lorry drivers & lorries |
# Question 3:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Volume, $X \sim N(32, 10^2)$ | | |
| $P(X < 40) = P\left(Z < \frac{40-32}{10}\right)$ | M1 | Standardising 40 with 32 and 10; allow $(32-40)$ |
| $= P(Z < 0.8)$ | A1 | CAO; ignore inequality and sign. May be implied by a correct answer |
| $= 0.788$ | A1 | AWRT $(0.78814)$ |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X > 25) = P(Z > -0.7)$ | | |
| $= P(Z < +0.7)$ | M1 | Area change. May be implied by a correct answer or an answer $> 0.5$ |
| $= 0.758$ | A1 | AWRT $(0.75804)$ |
## Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(25 < X < 40) = \text{(i)} - (1 - \text{(ii)})$ | M1 | OE; allow new start ignoring (i) & (ii). Allow even if incorrect standardising providing $0 <$ answer $< 1$. May be implied by a correct answer |
| $= 0.78814 - (1 - 0.75804) = 0.546$ | A1 | AWRT $(0.54618)$. Note: If (ii) is $0.242$, then $(0.788 - 0.242) = 0.546 \Rightarrow$ M0 A0 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B > \pounds 65) = P\left(Z > \frac{48.5-32}{10}\right)$ or $P\left(Z > \frac{65-42.88}{13.4}\right)$ | M1 | Attempt to change from $B$ to $X$ using $(48$ to $49)$, $32$ and $10$, or attempt to work with distribution of $B$ using $65$, $(42.8$ to $42.9)$ and $13.4$ |
| $= P(Z > 1.65) = 1 - P(Z < 1.65)$ | m1 | Area change. May be implied by a correct answer or an answer $< 0.5$ |
| $= 1 - 0.95053 = 0.049$ to $0.05(0)$ | A1 | AWFW $(0.04947)$ |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Other fuels; Other vehicles with an example (not other cars); Other types of customer; Minimum purchase (policy); Purchases in integer/fixed £s; Customers filling fuel cans | B2,1 | Size of car/engine/fuel tank $\Rightarrow$ B0; Price of fuel $\Rightarrow$ B0; Customer paying capacity $\Rightarrow$ B0. Must be two clearly different valid reasons for award of B2. Drivers and vehicles related $\Rightarrow$ B1, e.g. lorry drivers & lorries |
---3 During June 2011, the volume, $X$ litres, of unleaded petrol purchased per visit at a supermarket's filling station by private-car customers could be modelled by a normal distribution with a mean of 32 and a standard deviation of 10 .
\begin{enumerate}[label=(\alph*)]
\item Determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X < 40 )$;
\item $\mathrm { P } ( X > 25 )$;
\item $\mathrm { P } ( 25 < X < 40 )$.
\end{enumerate}\item Given that during June 2011 unleaded petrol cost $\pounds 1.34$ per litre, calculate the probability that the unleaded petrol bill for a visit during June 2011 by a private-car customer exceeded $\pounds 65$.
\item Give two reasons, in context, why the model $\mathrm { N } \left( 32,10 ^ { 2 } \right)$ is unlikely to be valid for a visit by any customer purchasing fuel at this filling station during June 2011.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2012 Q3 [12]}}