| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find parameters from quantiles |
| Difficulty | Standard +0.3 Part (a) is trivial arithmetic (IQR = Q3 - Q1). Part (b) requires knowing that quartiles correspond to z = ±0.674 for a normal distribution and solving simultaneous equations for μ and σ², which is a standard S1 technique but involves several steps. Part (c) is straightforward comparison. Overall slightly easier than average due to the routine nature of the calculations once the quartile-to-z-score relationship is recalled. |
| Spec | 2.02f Measures of average and spread2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(12.7 + 5.8 = 18.5\) minutes | A1 | |
| (b) \(P(X < 12.7) = 0.25\); \(P(Z < \frac{12.7-\mu}{\sigma}) = 0.25\) | M1 | |
| \(\frac{12.7-\mu}{\sigma} = 0.67\); \(12.7 - \mu = -0.67\sigma\) | M1 A1 | |
| \(P(X < 18.5) = 0.75\); \(P(Z < \frac{18.5-\mu}{\sigma}) = 0.75\) | M1 | |
| \(\frac{18.5-\mu}{\sigma} = 0.67\); \(18.5 - \mu = 0.67\sigma\) | M1 A1 | |
| Solve simultaneous giving \(\mu = 15.6\), \(\sigma = 4.3284\); so \(\mu = 15.6\), \(\sigma^2 = 18.7\) | M1 A1 | |
| (c) e.g. would expect normal dist. and mean and variance seem close to actual values so seems a fairly suitable model | B2 | (11 marks total) |
**(a)** $12.7 + 5.8 = 18.5$ minutes | A1 |
**(b)** $P(X < 12.7) = 0.25$; $P(Z < \frac{12.7-\mu}{\sigma}) = 0.25$ | M1 |
$\frac{12.7-\mu}{\sigma} = 0.67$; $12.7 - \mu = -0.67\sigma$ | M1 A1 |
$P(X < 18.5) = 0.75$; $P(Z < \frac{18.5-\mu}{\sigma}) = 0.75$ | M1 |
$\frac{18.5-\mu}{\sigma} = 0.67$; $18.5 - \mu = 0.67\sigma$ | M1 A1 |
Solve simultaneous giving $\mu = 15.6$, $\sigma = 4.3284$; so $\mu = 15.6$, $\sigma^2 = 18.7$ | M1 A1 |
**(c)** e.g. would expect normal dist. and mean and variance seem close to actual values so seems a fairly suitable model | B2 | **(11 marks total)**
---A call-centre dealing with complaints collected data on how long customers had to wait before an operator was free to take their call.
The lower quartile of the data was 12.7 minutes and the interquartile range was 5.8 minutes.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the upper quartile of the data. [1 mark]
\end{enumerate}
It is suggested that a normal distribution could be used to model the waiting time.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Calculate correct to 3 significant figures the mean and variance of this normal distribution based on the values of the quartiles. [8 marks]
\end{enumerate}
The actual mean and variance of the data were 15.3 minutes and 20.1 minutes$^2$ respectively.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Comment on the suitability of the model. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q3 [11]}}