| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.8 This is a straightforward S1 normal distribution question testing standard procedures: standardizing to find probabilities using tables, combining probabilities with basic set operations, and working backwards from a probability to find a parameter. All parts follow routine algorithms with no problem-solving insight required, making it easier than average but not trivial due to the multiple parts and inverse normal calculation in part (b). |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (i) \(P(X < 1.90) = P\!\left(Z < \frac{1.90-1.86}{0.04}\right) = P(Z < 1) = 0.8413\) | M1 A1 | |
| (ii) \(P(X > 1.80) = P\!\left(Z > \frac{1.80-1.86}{0.04}\right) = P(Z > -1.5) = 0.9332\) | M1 A1 | |
| (iii) \(P(1.80 < X < 1.90) = 0.9332 + 0.8413 - 1 = 0.7745\) | M1 A1 | Follow through from (i) and (ii) |
| (iv) \(P(X \neq 1.86) = 1\) | B1 | Since normal distribution is continuous |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 1.80) = 0.98 \Rightarrow P\!\left(Z > \frac{1.80 - 1.86}{\sigma}\right) = 0.98\) | M1 | |
| \(\frac{1.80 - 1.86}{\sigma} = -2.054\) | A1 | Using \(z = -2.054\) (or \(\pm 2.054\)) |
| \(\sigma = \frac{0.06}{2.054} \approx 0.0292\) | A1 |
# Question 2:
## Part (a) - $X \sim N(1.86, 0.04^2)$
| Answer | Mark | Guidance |
|--------|------|----------|
| (i) $P(X < 1.90) = P\!\left(Z < \frac{1.90-1.86}{0.04}\right) = P(Z < 1) = 0.8413$ | M1 A1 | |
| (ii) $P(X > 1.80) = P\!\left(Z > \frac{1.80-1.86}{0.04}\right) = P(Z > -1.5) = 0.9332$ | M1 A1 | |
| (iii) $P(1.80 < X < 1.90) = 0.9332 + 0.8413 - 1 = 0.7745$ | M1 A1 | Follow through from (i) and (ii) |
| (iv) $P(X \neq 1.86) = 1$ | B1 | Since normal distribution is continuous |
## Part (b) - Finding $\sigma$
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 1.80) = 0.98 \Rightarrow P\!\left(Z > \frac{1.80 - 1.86}{\sigma}\right) = 0.98$ | M1 | |
| $\frac{1.80 - 1.86}{\sigma} = -2.054$ | A1 | Using $z = -2.054$ (or $\pm 2.054$) |
| $\sigma = \frac{0.06}{2.054} \approx 0.0292$ | A1 | |2 A garden centre sells bamboo canes of nominal length 1.8 metres. The length, $X$ metres, of the canes can be modelled by a normal distribution with mean 1.86 and standard deviation $\sigma$.
\begin{enumerate}[label=(\alph*)]
\item Assuming that $\sigma = 0.04$, determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X < 1.90 )$;
\item $\mathrm { P } ( X > 1.80 )$;
\item $\mathrm { P } ( 1.80 < X < 1.90 )$;
\item $\mathrm { P } ( X \neq 1.86 )$.
\end{enumerate}\item It is subsequently found that $\mathrm { P } ( X > 1.80 ) = 0.98$.
Determine the value of $\sigma$.\\[0pt]
[3 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-06_1529_1717_1178_150}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2014 Q2 [10]}}