| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | March |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring inverse normal calculations to find μ and σ from percentiles, then a binomial probability calculation, and finally a straightforward normal probability with a simple relationship between parameters. All techniques are routine for this level with no novel problem-solving required, making it slightly easier than average. |
| 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 |
|---|---|---|
| \(0.674 = \dfrac{8.8 - \mu}{\sigma} \Rightarrow 0.674\sigma = 8.8 - \mu\) | B1 | \(\pm 0.674\) seen |
| \(-0.935 = \dfrac{7.7 - \mu}{\sigma} \Rightarrow -0.935\sigma = 7.7 - \mu\) | B1 | \(\pm 0.935\) seen (condone \(\pm 0.934\)) |
| M1 | An eqn with a \(z\)-value, \(\mu\) and \(\sigma\); allow sq rt, sq cc | |
| M1 | sensible attempt to eliminate \(\mu\) or \(\sigma\) by substitution or subtraction | |
| \(\sigma = 0.684\), \(\mu = 8.34\) | A1 | correct answers (from \(-0.935\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(<8.2) = P\!\left(z < \dfrac{8.2 - 7.9}{0.44}\right)\) | M1 | Standardising no cc no sq rt no sq |
| M1 | Correct area ie \(\Phi\), final solution | |
| \(= P(z < 0.6818) = 0.7524\) | A1 | Correct prob rounding to \(0.752\) |
| \(P(3) = {}^{5}C_3\ (0.7524)^3(0.2476)^2\) | M1 | Binomial \({}^5C_x\), powers summing to 5, any \(p\), \(\Sigma p = 1\) |
| \(= 0.261\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(<1.5\mu) = P\!\left(z < \dfrac{1.5\mu - \mu}{\mu}\right) = P(z < 0.5)\) | *M1 | standardising with \(\mu\) and \(\sigma\) (\(\sigma\) may be replaced by \(\mu\)) |
| DM1 | just one variable | |
| \(= 0.692\) | A1 |
# Question 7(a)(i):
$0.674 = \dfrac{8.8 - \mu}{\sigma} \Rightarrow 0.674\sigma = 8.8 - \mu$ | B1 | $\pm 0.674$ seen
$-0.935 = \dfrac{7.7 - \mu}{\sigma} \Rightarrow -0.935\sigma = 7.7 - \mu$ | B1 | $\pm 0.935$ seen (condone $\pm 0.934$)
| M1 | An eqn with a $z$-value, $\mu$ and $\sigma$; allow sq rt, sq cc
| M1 | sensible attempt to eliminate $\mu$ or $\sigma$ by substitution or subtraction
$\sigma = 0.684$, $\mu = 8.34$ | A1 | correct answers (from $-0.935$)
**Total: 5**
---
# Question 7(a)(ii):
$P(<8.2) = P\!\left(z < \dfrac{8.2 - 7.9}{0.44}\right)$ | M1 | Standardising no cc no sq rt no sq
| M1 | Correct area ie $\Phi$, final solution
$= P(z < 0.6818) = 0.7524$ | A1 | Correct prob rounding to $0.752$
$P(3) = {}^{5}C_3\ (0.7524)^3(0.2476)^2$ | M1 | Binomial ${}^5C_x$, powers summing to 5, any $p$, $\Sigma p = 1$
$= 0.261$ | A1 |
**Total: 5**
---
# Question 7(b):
$P(<1.5\mu) = P\!\left(z < \dfrac{1.5\mu - \mu}{\mu}\right) = P(z < 0.5)$ | *M1 | standardising with $\mu$ and $\sigma$ ($\sigma$ may be replaced by $\mu$)
| DM1 | just one variable
$= 0.692$ | A1 |
**Total: 3**7
\begin{enumerate}[label=(\alph*)]
\item The lengths, in centimetres, of middle fingers of women in Raneland have a normal distribution with mean $\mu$ and standard deviation $\sigma$. It is found that $25 \%$ of these women have fingers longer than 8.8 cm and $17.5 \%$ have fingers shorter than 7.7 cm .
\begin{enumerate}[label=(\roman*)]
\item Find the values of $\mu$ and $\sigma$.\\
The lengths, in centimetres, of middle fingers of women in Snoland have a normal distribution with mean 7.9 and standard deviation 0.44. A random sample of 5 women from Snoland is chosen.
\item Find the probability that exactly 3 of these women have middle fingers shorter than 8.2 cm .
\end{enumerate}\item The random variable $X$ has a normal distribution with mean equal to the standard deviation. Find the probability that a particular value of $X$ is less than 1.5 times the mean.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2017 Q7 [13]}}