| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct comparison of probabilities |
| Difficulty | Moderate -0.8 This is a straightforward S1 normal distribution question requiring standard z-score calculations and inverse normal lookup. Part (a) involves routine probability calculations (including the conceptual point that P(X = exactly 90) = 0 for continuous distributions), while part (b) requires working backwards from a percentile to find σ using tables/calculator. All techniques are standard textbook exercises with no novel problem-solving required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim N(91, 0.8^2)\), \(P(X < 90) = P\left(Z < \frac{90-91}{0.8}\right) = P(Z < -1.25)\) | M1 | Standardising with 91 and 0.8 |
| \(= 1 - \Phi(1.25) = 1 - 0.8944\) | A1 | |
| \(= \mathbf{0.1056}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Since \(X\) is continuous, \(P(X = 90) = 0\), so \(P(\text{not exactly } 90) = \mathbf{1}\) | B1 | Must state continuous distribution or equivalent reasoning |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(91 < X < 92.5) = P\left(0 < Z < \frac{92.5-91}{0.8}\right) = P(0 < Z < 1.875)\) | M1 | Standardising correctly |
| \(= \Phi(1.875) - 0.5\) | M1 | Correct method using symmetry/tables |
| \(= 0.9696 - 0.5 = \mathbf{0.4696}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 150) = 0.01\), so \(P\left(Z < \frac{150-153}{\sigma}\right) = 0.01\) | M1 | Setting up correct probability statement |
| \(\frac{150-153}{\sigma} = -2.3263\) | M1 | Using \(z = -2.3263\) (or \(\pm\)2.326) |
| \(\sigma = \frac{3}{2.3263}\) | A1 | |
| \(\sigma = \mathbf{1.29}\) metres | A1 | cao, accept 1.28–1.29 |
# Question 2:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim N(91, 0.8^2)$, $P(X < 90) = P\left(Z < \frac{90-91}{0.8}\right) = P(Z < -1.25)$ | M1 | Standardising with 91 and 0.8 |
| $= 1 - \Phi(1.25) = 1 - 0.8944$ | A1 | |
| $= \mathbf{0.1056}$ | A1 | |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Since $X$ is continuous, $P(X = 90) = 0$, so $P(\text{not exactly } 90) = \mathbf{1}$ | B1 | Must state continuous distribution or equivalent reasoning |
## Part (a)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(91 < X < 92.5) = P\left(0 < Z < \frac{92.5-91}{0.8}\right) = P(0 < Z < 1.875)$ | M1 | Standardising correctly |
| $= \Phi(1.875) - 0.5$ | M1 | Correct method using symmetry/tables |
| $= 0.9696 - 0.5 = \mathbf{0.4696}$ | A1 | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 150) = 0.01$, so $P\left(Z < \frac{150-153}{\sigma}\right) = 0.01$ | M1 | Setting up correct probability statement |
| $\frac{150-153}{\sigma} = -2.3263$ | M1 | Using $z = -2.3263$ (or $\pm$2.326) |
| $\sigma = \frac{3}{2.3263}$ | A1 | |
| $\sigma = \mathbf{1.29}$ metres | A1 | cao, accept 1.28–1.29 |
---2 The length of aluminium baking foil on a roll may be modelled by a normal distribution with mean 91 metres and standard deviation 0.8 metres.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that the length of foil on a particular roll is:
\begin{enumerate}[label=(\roman*)]
\item less than 90 metres;
\item not exactly 90 metres;
\item between 91 metres and 92.5 metres.
\end{enumerate}\item The length of cling film on a roll may also be modelled by a normal distribution but with mean 153 metres and standard deviation $\sigma$ metres.
It is required that $1 \%$ of rolls of cling film should have a length less than 150 metres.\\
Find the value of $\sigma$ that is needed to satisfy this requirement.\\[0pt]
[4 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{4c679380-894f-4d36-aec8-296b662058e2-04_1526_1714_1181_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q2 [10]}}