| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Expected frequency with unknown parameter |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution problem requiring inverse normal calculations to find parameters from given probabilities, then a straightforward expected frequency calculation. While it involves multiple steps, the techniques are routine and commonly practiced at this level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{15-\mu}{\sigma}=1.2816\) | M1 | \(\pm\left(\frac{15-\mu}{\sigma}\right)=(z\text{ value})\), \(1<\ |
| \(\dfrac{5-\mu}{\sigma}=-1.6449\) | M1 B1 | \(\pm\left(\frac{5-\mu}{\sigma}\right)=(z\text{ value})\), \(\ |
| \(10=2.9265\sigma\) | depM1 | Solving simultaneous equations; dependent on at least 1 previous M mark |
| \(\sigma=3.41705\ldots\) awrt \(\mathbf{3.42}\) | A1 | Note: use of 0.8159 and 0.5199 as z-values scores 0/6 |
| \(\mu=10.6207\ldots\) awrt \(\mathbf{10.6}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Number of people \(=[23\times]P(L>12)\) \(=[23\times]P\!\left(Z>\dfrac{12-10.6}{3.42}\right)\) | M1 | M1 for \(P\!\left(Z>\frac{12-\text{their }10.6}{\text{their }3.42}\right)\), only ft if \(\sigma>0\) |
| \(=23\times(1-0.6591)\) | depM1 | dep on 1st M1; for \(23\times\text{their}\,P(L>12)\) where \(0 |
| \(\approx\) awrt \(7.8\)/awrt \(7.9\) | A1 | Allow 7 or 8 from correct working |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{15-\mu}{\sigma}=1.2816$ | M1 | $\pm\left(\frac{15-\mu}{\sigma}\right)=(z\text{ value})$, $1<\|z\|<1.5$ |
| $\dfrac{5-\mu}{\sigma}=-1.6449$ | M1 B1 | $\pm\left(\frac{5-\mu}{\sigma}\right)=(z\text{ value})$, $\|z\|>1.5$; B1 both $\pm1.2816$ and $\pm1.6449$ correct |
| $10=2.9265\sigma$ | depM1 | Solving simultaneous equations; dependent on at least 1 previous M mark |
| $\sigma=3.41705\ldots$ awrt $\mathbf{3.42}$ | A1 | Note: use of 0.8159 and 0.5199 as z-values scores 0/6 |
| $\mu=10.6207\ldots$ awrt $\mathbf{10.6}$ | A1 | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Number of people $=[23\times]P(L>12)$ $=[23\times]P\!\left(Z>\dfrac{12-10.6}{3.42}\right)$ | M1 | M1 for $P\!\left(Z>\frac{12-\text{their }10.6}{\text{their }3.42}\right)$, only ft if $\sigma>0$ |
| $=23\times(1-0.6591)$ | depM1 | dep on 1st M1; for $23\times\text{their}\,P(L>12)$ where $0<P(L>12)<0.5$; A1 must come from correct $\mu$ and $\sigma=$ awrt 3.4 |
| $\approx$ awrt $7.8$/awrt $7.9$ | A1 | Allow 7 or 8 from correct working |
---6. The waiting time, $L$ minutes, to see a doctor at a health centre is normally distributed with $L \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$.
Given that $\mathrm { P } ( L < 15 ) = 0.9$ and $\mathrm { P } ( L < 5 ) = 0.05$
\begin{enumerate}[label=(\alph*)]
\item find the value of $\mu$ and the value of $\sigma$.
There are 23 people waiting to see a doctor at the health centre.
\item Determine the expected number of these people who will have a waiting time of more than 12 minutes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2018 Q6 [9]}}