| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Expected frequency with unknown parameter |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: z-score calculations, inverse normal to find an unknown mean, and binomial probability. Part (a)(ii) involves working backwards from a probability to find a parameter, which is slightly beyond pure recall, but all parts use routine S1 methods with no novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(x > 3900) = P\!\left(z > \frac{3900 - 4520}{560}\right)\) | M1 | Standardising, no cc, no sq rt, no sq |
| \(= P(z > -1.107) = \Phi(1.107) = 0.8657\) | M1 | Correct area \(\Phi\) i.e. \(> 0.5\) |
| Number of days \(= 365 \times 0.8657\) | A1 | Prob rounding to 0.866 |
| \(= 315\) or \(316\ (315.98)\) | B1\(\checkmark\) (4) | Correct answer ft their wrong prob if previous A0, \(p < 1\), ft must be accurate to 3sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z = 1.165\) | B1 | \(\pm 1.165\) seen |
| \(1.165 = \frac{8000 - m}{560}\) | M1 | Standardising eqn, allow sq, sq rt, cc, must have \(z\)-value e.g. not 0.122, 0.878, 0.549, 0.810 |
| \(m = 7350\ (7347.6)\) | A1 (3) | Correct answer rounding to 7350 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(0,1) = (0.878)^6 + {^6C_1}(0.122)^1(0.878)^5\) | M1 | Binomial term \(^6C_x\, p^x(1-p)^{6-x}\), \(0 < p < 1\) seen |
| M1 | Correct unsimplified expression | |
| \(= 0.840\) accept \(0.84\). Normal approx to Binomial: M0, M0, A0 | A1 (3) | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(< 2\mu) = P\!\left(z > \frac{2\mu - \mu}{\sigma}\right) = P(z < 1.5)\) | M1 | Standardising with \(\mu\) and \(\sigma\) |
| M1 | Attempt at one variable and cancel | |
| \(= 0.933\) | A1 (3) | Correct answer |
## Question 7:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(x > 3900) = P\!\left(z > \frac{3900 - 4520}{560}\right)$ | M1 | Standardising, no cc, no sq rt, no sq |
| $= P(z > -1.107) = \Phi(1.107) = 0.8657$ | M1 | Correct area $\Phi$ i.e. $> 0.5$ |
| Number of days $= 365 \times 0.8657$ | A1 | Prob rounding to 0.866 |
| $= 315$ or $316\ (315.98)$ | B1$\checkmark$ (4) | Correct answer ft their wrong prob if previous A0, $p < 1$, ft must be accurate to 3sf |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = 1.165$ | B1 | $\pm 1.165$ seen |
| $1.165 = \frac{8000 - m}{560}$ | M1 | Standardising eqn, allow sq, sq rt, cc, must have $z$-value e.g. not 0.122, 0.878, 0.549, 0.810 |
| $m = 7350\ (7347.6)$ | A1 (3) | Correct answer rounding to 7350 |
### Part (a)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(0,1) = (0.878)^6 + {^6C_1}(0.122)^1(0.878)^5$ | M1 | Binomial term $^6C_x\, p^x(1-p)^{6-x}$, $0 < p < 1$ seen |
| | M1 | Correct unsimplified expression |
| $= 0.840$ accept $0.84$. Normal approx to Binomial: M0, M0, A0 | A1 (3) | Correct answer |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(< 2\mu) = P\!\left(z > \frac{2\mu - \mu}{\sigma}\right) = P(z < 1.5)$ | M1 | Standardising with $\mu$ and $\sigma$ |
| | M1 | Attempt at one variable and cancel |
| $= 0.933$ | A1 (3) | Correct answer |7
\begin{enumerate}[label=(\alph*)]
\item A petrol station finds that its daily sales, in litres, are normally distributed with mean 4520 and standard deviation 560.
\begin{enumerate}[label=(\roman*)]
\item Find on how many days of the year ( 365 days) the daily sales can be expected to exceed 3900 litres.
The daily sales at another petrol station are $X$ litres, where $X$ is normally distributed with mean $m$ and standard deviation 560. It is given that $\mathrm { P } ( X > 8000 ) = 0.122$.
\item Find the value of $m$.
\item Find the probability that daily sales at this petrol station exceed 8000 litres on fewer than 2 of 6 randomly chosen days.
\end{enumerate}\item The random variable $Y$ is normally distributed with mean $\mu$ and standard deviation $\sigma$. Given that $\sigma = \frac { 2 } { 3 } \mu$, find the probability that a random value of $Y$ is less than $2 \mu$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2015 Q7 [13]}}