| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Validity of normal model |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard procedures: sketching, finding probabilities using tables/calculator, and inverse normal calculations. All parts follow routine S1 techniques with no conceptual challenges, though part (d) requires careful handling of conditional probability regions. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sketch of symmetric bell-shaped curve not crossing \(x\)-axis, with 55 labelled at centre | B1 dB1 | 1st B1 for reasonable symmetric bell-shaped curve not crossing \(x\)-axis (ignore any vertical axis drawn); 2nd B1 dependent on previous B1 for 55 labelled at centre of \(x\)-axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X>70) = P\!\left(Z > \frac{70-55}{20}\right) = P(Z > 0.75)\) | M1 | 1st M1 for standardising with 70, 55, 20 (allow \(\pm\)) |
| \(= 1 - 0.7734 = 0.2266\), awrt 0.227/22.7% | M1A1 | 2nd M1 use of \(1-p\) (must be a probability, so \(1-0.67\) is M0); A1 awrt 0.227 or 22.7% |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > b) = 0.01\) | ||
| \(\frac{b-55}{20} = 2.3263\) | M1B1 | M1 for standardising with 55, 20 and equating to \(z\)-value \(\lvert z\rvert > 2\); B1 for 2.3263 (or better) used and compatible sign with their standardisation |
| \(b = 101.526\), Given answer 102 | A1 | A1 for awrt 102 which must come from a \(z\)-value in range \(2.32 \leq z \leq 2.34\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(70 < X < m) = 0.1315\) \(P(X < m) - P(X < 70) = 0.1315\) | M1 | 1st M1 for correct expression for \(P(X |
| \(P\!\left(Z < \frac{m-55}{20}\right) = 0.9049\) | ||
| \(\frac{m-55}{20} = 1.31\) | M1B1 | 2nd M1 for standardising with 55, 20 and equating to \(z\)-value \(\lvert z\rvert > 1\); B1 1.31 (1.31018… from calc) used and compatible sign with their standardisation |
| \(m = 81.2\) | A1 | awrt 81.2 |
## Question 5:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch of symmetric bell-shaped curve not crossing $x$-axis, with 55 labelled at centre | B1 dB1 | 1st B1 for reasonable symmetric bell-shaped curve not crossing $x$-axis (ignore any vertical axis drawn); 2nd B1 dependent on previous B1 for 55 labelled at centre of $x$-axis |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X>70) = P\!\left(Z > \frac{70-55}{20}\right) = P(Z > 0.75)$ | M1 | 1st M1 for standardising with 70, 55, 20 (allow $\pm$) |
| $= 1 - 0.7734 = 0.2266$, awrt **0.227/22.7%** | M1A1 | 2nd M1 use of $1-p$ (must be a probability, so $1-0.67$ is M0); A1 awrt 0.227 or 22.7% |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > b) = 0.01$ | | |
| $\frac{b-55}{20} = 2.3263$ | M1B1 | M1 for standardising with 55, 20 and equating to $z$-value $\lvert z\rvert > 2$; B1 for 2.3263 (or better) used and compatible sign with their standardisation |
| $b = 101.526$, **Given answer 102** | A1 | A1 for awrt 102 which must come from a $z$-value in range $2.32 \leq z \leq 2.34$ |
### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(70 < X < m) = 0.1315$ $P(X < m) - P(X < 70) = 0.1315$ | M1 | 1st M1 for correct expression for $P(X<m)$ e.g. $0.1315 + \text{'0.7734'}$ or $P(X>m)=0.0951$ or sight of 0.9049 (may be implied by sight of 1.31) |
| $P\!\left(Z < \frac{m-55}{20}\right) = 0.9049$ | | |
| $\frac{m-55}{20} = 1.31$ | M1B1 | 2nd M1 for standardising with 55, 20 and equating to $z$-value $\lvert z\rvert > 1$; B1 1.31 (1.31018… from calc) used and compatible sign with their standardisation |
| $m = 81.2$ | A1 | awrt 81.2 |\begin{enumerate}
\item Police measure the speed of cars passing a particular point on a motorway. The random variable $X$ is the speed of a car.\\
$X$ is modelled by a normal distribution with mean 55 mph (miles per hour).\\
(a) Draw a sketch to illustrate the distribution of $X$. Label the mean on your sketch.
\end{enumerate}
The speed limit on the motorway is 70 mph . Car drivers can choose to travel faster than the speed limit but risk being caught by the police.
The distribution of $X$ has a standard deviation of 20 mph .\\
(b) Find the percentage of cars that are travelling faster than the speed limit.
The fastest $1 \%$ of car drivers will be banned from driving.\\
(c) Show that the lowest speed, correct to 3 significant figures, for a car driver to be banned is 102 mph . Show your working clearly.
Car drivers will just be given a caution if they are travelling at a speed $m$ such that
$$\mathrm { P } ( 70 < X < m ) = 0.1315$$
(d) Find the value of $m$. Show your working clearly.
\hfill \mbox{\textit{Edexcel S1 2015 Q5 [12]}}