| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution problem requiring inverse normal calculations to find parameters from percentiles, then using those parameters to check model validity. While it involves multiple steps and careful handling of percentiles/z-scores, it follows a routine template taught in S1 with no novel problem-solving required. The 9 marks reflect length rather than conceptual difficulty, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | M2 | \(\mathrm{P}(X > 2) = 0.75\); \(\mathrm{P}(Z > \frac{2-\mu}{\sigma}) = 0.75\) |
| A1 | \(\frac{2-\mu}{\sigma} = -0.67\); \(2 - \mu = -0.67\sigma\) | |
| M2 | \(\mathrm{P}(X > 6) = 0.1\); \(\mathrm{P}(Z > \frac{6-\mu}{\sigma}) = 0.1\) | |
| A1 | \(\frac{6-\mu}{\sigma} = 1.2816\); \(6 - \mu = 1.2816\sigma\) | |
| M1 A2 | Solve simultaneously giving \(\mu = 3.3732, \sigma = 2.0496\); so \(\mu = 3.37, \sigma = 2.05\) | |
| (b) | M2 A1 | \(\mathrm{P}(X < 1) = \mathrm{P}(Z < \frac{1-3.3732}{2.0496}) = \mathrm{P}(Z < -1.16) = 0.1230 \therefore 12.3\%\) |
| (c) | B2, (14) | e.g. large discrepancy between predicted & actual \(\therefore\) not v. suitable |
(a) | M2 | $\mathrm{P}(X > 2) = 0.75$; $\mathrm{P}(Z > \frac{2-\mu}{\sigma}) = 0.75$
| A1 | $\frac{2-\mu}{\sigma} = -0.67$; $2 - \mu = -0.67\sigma$
| M2 | $\mathrm{P}(X > 6) = 0.1$; $\mathrm{P}(Z > \frac{6-\mu}{\sigma}) = 0.1$
| A1 | $\frac{6-\mu}{\sigma} = 1.2816$; $6 - \mu = 1.2816\sigma$
| M1 A2 | Solve simultaneously giving $\mu = 3.3732, \sigma = 2.0496$; so $\mu = 3.37, \sigma = 2.05$
(b) | M2 A1 | $\mathrm{P}(X < 1) = \mathrm{P}(Z < \frac{1-3.3732}{2.0496}) = \mathrm{P}(Z < -1.16) = 0.1230 \therefore 12.3\%$
(c) | B2, (14) | e.g. large discrepancy between predicted & actual $\therefore$ not v. suitable
---A geologist is analysing the size of quartz crystals in a sample of granite. She estimates that the longest diameter of 75% of the crystals is greater than 2 mm, but only 10% of the crystals have a longest diameter of more than 6 mm.
The geologist believes that the distribution of the longest diameters of the quartz crystals can be modelled by a normal distribution.
\begin{enumerate}[label=(\alph*)]
\item Find the mean and variance of this normal distribution. [9 marks]
\end{enumerate}
The geologist also estimated that only 2% of the longest diameters were smaller than 1 mm.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Calculate the corresponding percentage that would be predicted by a normal distribution with the parameters you calculated in part $(a)$. [3 marks]
\item Hence, comment on the suitability of the normal distribution as a model in this situation. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [14]}}