| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| 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 only standard table lookups and z-score calculations. Part (a) involves three routine probability calculations with given parameters, and part (b) is a standard inverse normal problem using the 80th percentile. All techniques are direct applications of formulas with no problem-solving insight required, making it easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\left(Z \leq \frac{10-7.5}{1.6}\right) = P(Z \leq 1.5625)\) | M1 | Standardising |
| \(= 0.9410\) | A1 | Accept 0.941 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\left(Z > \frac{6-7.5}{1.6}\right) = P(Z > -0.9375)\) | M1 | Standardising |
| \(= P(Z < 0.9375) = 0.8257\) | A1 | Accept 0.826 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\left(\frac{5-7.5}{1.6} < Z < \frac{10-7.5}{1.6}\right) = P(-1.5625 < Z < 1.5625)\) | M1 | Correct method combining probabilities |
| \(= 2(0.9410) - 1 = 0.8821\) | A1 | Accept 0.882 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 15) = 0.80\) | B1 | Correct probability statement |
| \(\frac{15 - \mu}{2.4} = 0.8416\) | M1 | Using \(z = 0.8416\) (from tables) |
| \(15 - \mu = 2.4 \times 0.8416\) | M1 | Correct rearrangement |
| \(\mu = 15 - 2.019 = 12.98\) (awrt \(13.0\)) | A1 | cao |
# Question 2:
## Part (a) - Normal Distribution, $X \sim N(7.5, 1.6^2)$ [6 marks]
**(i) P(X ≤ 10)**
| $P\left(Z \leq \frac{10-7.5}{1.6}\right) = P(Z \leq 1.5625)$ | M1 | Standardising |
|---|---|---|
| $= 0.9410$ | A1 | Accept 0.941 |
**(ii) P(X > 6)**
| $P\left(Z > \frac{6-7.5}{1.6}\right) = P(Z > -0.9375)$ | M1 | Standardising |
|---|---|---|
| $= P(Z < 0.9375) = 0.8257$ | A1 | Accept 0.826 |
**(iii) P(5 < X < 10)**
| $P\left(\frac{5-7.5}{1.6} < Z < \frac{10-7.5}{1.6}\right) = P(-1.5625 < Z < 1.5625)$ | M1 | Correct method combining probabilities |
|---|---|---|
| $= 2(0.9410) - 1 = 0.8821$ | A1 | Accept 0.882 |
## Part (b) - Finding $\mu$, $X \sim N(\mu, 2.4^2)$ [4 marks]
| $P(X < 15) = 0.80$ | B1 | Correct probability statement |
|---|---|---|
| $\frac{15 - \mu}{2.4} = 0.8416$ | M1 | Using $z = 0.8416$ (from tables) |
| $15 - \mu = 2.4 \times 0.8416$ | M1 | Correct rearrangement |
| $\mu = 15 - 2.019 = 12.98$ (awrt $13.0$) | A1 | cao |2
\begin{enumerate}[label=(\alph*)]
\item Tim rings the church bell in his village every Sunday morning. The time that he spends ringing the bell may be modelled by a normal distribution with mean 7.5 minutes and standard deviation 1.6 minutes.
Determine the probability that, on a particular Sunday morning, the time that Tim spends ringing the bell is:
\begin{enumerate}[label=(\roman*)]
\item at most 10 minutes;
\item more than 6 minutes;
\item between 5 minutes and 10 minutes.
\end{enumerate}\item June rings the same church bell for weekday weddings. The time that she spends, in minutes, ringing the bell may be modelled by the distribution $\mathrm { N } \left( \mu , 2.4 ^ { 2 } \right)$.
Given that 80 per cent of the times that she spends ringing the bell are less than 15 minutes, find the value of $\mu$.\\[0pt]
[4 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{ddf7f158-b6ae-42c6-98f1-d59c205646ad-04_1477_1707_1226_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2014 Q2 [10]}}