| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear relationship μ = kσ |
| Difficulty | Standard +0.3 Parts (a)(i) and (a)(ii) are routine normal distribution calculations using tables or inverse normal. Part (b) requires recognizing that P(X > 0) = P(Z > -μ/σ) = P(Z > -3/2), which is a straightforward standardization with the given linear relationship. This is slightly above average difficulty due to the algebraic manipulation in part (b), but remains a standard S1 question type. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X < 170) =] P\left(Z < \frac{170-166}{10}\right)\) | M1 | Use of \(\pm\) standardisation formula with 170, 166 and 10 substituted appropriately; condone \(10^2\), \(\sqrt{10}\), condone continuity correction \(\pm 0.5\) |
| \([= P(Z < 0.4) =]\ 0.655\) | A1 | \(0.655 \leqslant p < 0.6555\); If M0 awarded, SC B1 for correct answer www |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[P\left(Z > \frac{h-166}{10}\right) = 0.4\right]\) | B1 | \(0.253 \leqslant z \leqslant 0.2535\) or \(-0.2535 \leqslant z \leqslant -0.253\) seen |
| \(\frac{h-166}{10} = 0.253\) | M1 | Use of \(\pm\) standardisation formula with \(h\), 166, 10 and a \(z\)-value (not \(1-z\)-value); not \(10^2\), \(\sqrt{10}\), no continuity correction |
| \(h = 168.53\) | A1 | If M0 scored, SC B1 for \(168.53 \leqslant h \leqslant 168.535\), 168.5; SC B1 for 168.54 from \(z = 0.254\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[P(X>0) = P\left(Z > \frac{0-\mu}{\sigma}\right) =\right] P\left(Z > \frac{[0]-\mu}{\frac{2}{3}\mu}\right)\) or \(P\left(Z > \frac{[0]-\frac{3}{2}\sigma}{\sigma}\right)\) | M1 | Use of \(\pm\) standardisation formula with 0, \(\mu\) and \(\frac{2}{3}\mu\) substituted for \(\sigma\); or use of \(\pm\) standardisation formula with 0, \(\sigma\) and \(\frac{3}{2}\sigma\) substituted for \(\mu\) |
| \(= P(Z > -1.5)\) | A1 | \(-1.5\) seen, no additional terms (e.g. \(x - 1.5\), A0); condone \(Z < 1.5\); If M0 scored, SC B1 for \(Z > -1.5\) or \(Z < 1.5\) seen www |
| \(= 0.933\) final answer | A1 | \(0.933 \leqslant p < 0.9333\); If M0 scored, SC B1 for \(0.933 \leqslant p < 0.9333\) seen www |
## Question 5(a)(i):
$[P(X < 170) =] P\left(Z < \frac{170-166}{10}\right)$ | M1 | Use of $\pm$ standardisation formula with 170, 166 and 10 substituted appropriately; condone $10^2$, $\sqrt{10}$, condone continuity correction $\pm 0.5$
$[= P(Z < 0.4) =]\ 0.655$ | A1 | $0.655 \leqslant p < 0.6555$; If M0 awarded, SC B1 for correct answer www
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## Question 5(a)(ii):
$\left[P\left(Z > \frac{h-166}{10}\right) = 0.4\right]$ | B1 | $0.253 \leqslant z \leqslant 0.2535$ or $-0.2535 \leqslant z \leqslant -0.253$ seen
$\frac{h-166}{10} = 0.253$ | M1 | Use of $\pm$ standardisation formula with $h$, 166, 10 and a $z$-value (not $1-z$-value); not $10^2$, $\sqrt{10}$, no continuity correction
$h = 168.53$ | A1 | If M0 scored, SC B1 for $168.53 \leqslant h \leqslant 168.535$, 168.5; SC B1 for 168.54 from $z = 0.254$
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## Question 5(b):
$\left[P(X>0) = P\left(Z > \frac{0-\mu}{\sigma}\right) =\right] P\left(Z > \frac{[0]-\mu}{\frac{2}{3}\mu}\right)$ or $P\left(Z > \frac{[0]-\frac{3}{2}\sigma}{\sigma}\right)$ | M1 | Use of $\pm$ standardisation formula with 0, $\mu$ and $\frac{2}{3}\mu$ substituted for $\sigma$; or use of $\pm$ standardisation formula with 0, $\sigma$ and $\frac{3}{2}\sigma$ substituted for $\mu$
$= P(Z > -1.5)$ | A1 | $-1.5$ seen, no additional terms (e.g. $x - 1.5$, A0); condone $Z < 1.5$; If M0 scored, SC B1 for $Z > -1.5$ or $Z < 1.5$ seen www
$= 0.933$ final answer | A1 | $0.933 \leqslant p < 0.9333$; If M0 scored, SC B1 for $0.933 \leqslant p < 0.9333$ seen www
---5
\begin{enumerate}[label=(\alph*)]
\item The heights of the members of a club are normally distributed with mean 166 cm and standard deviation 10 cm .
\begin{enumerate}[label=(\roman*)]
\item Find the probability that a randomly chosen member of the club has height less than 170 cm .
\item Given that $40 \%$ of the members have heights greater than $h \mathrm {~cm}$, find the value of $h$ correct to 2 decimal places.
\end{enumerate}\item The random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$. Given that $\sigma = \frac { 2 } { 3 } \mu$, find the probability that a randomly chosen value of $X$ is positive.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q5 [8]}}