| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Combined events across distributions |
| Difficulty | Standard +0.3 This is a straightforward normal distribution question requiring standard techniques: z-score calculations for probability (part a), inverse normal to find standard deviation (part b), and combining independent probabilities (part c). All parts are routine applications with no novel problem-solving required, 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(46 < X < 62) = P\!\left(\frac{46-55}{6} < Z < \frac{62-55}{6}\right)\) | M1 | 46 or 62, 55 and 6 substituted into \(\pm\)standardisation formula once. Condone \(6^2\) and continuity correction \(\pm 0.5\) |
| \(= P\!\left(-1.5 < Z < \frac{7}{6}\right)\) | B1 | Both standardisation values correct, accept unsimplified |
| \(\left[=\Phi\!\left(\frac{7}{6}\right) - (1 - \Phi(1.5))\right] = 0.8784 + (0.9332-1)\) | M1 | Calculating the appropriate area from stated \(\Phi\)s of \(z\)-values, must be probabilities |
| \(0.812\) | A1 | \(0.8115 < p \leqslant 0.812\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = \pm 0.674\) | B1 | CAO, critical \(z\)-value |
| \(\frac{36-42}{\sigma} = -0.674\) | M1 | 36 and 42 substituted in \(\pm\)standardisation formula, no continuity correction, not \(\sigma^2\), \(\sqrt{\sigma}\), equated to a \(z\)-value |
| \(\sigma = 8.9[0]\) | A1 | WWW. Only dependent on M |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{male} < 46) = 1 - \textit{their}\ 0.9332 = 0.0668\) | M1 | FT value from part (a) or Correct: \(1 - \Phi\!\left(\frac{46-55}{6}\right)\), condone continuity correction, \(\sigma^2\), \(\sqrt{\sigma}\), and probability found. Condone unsupported correct value stated |
| \(P(\text{female} < 46) = P\!\left(Z < \frac{46-42}{\textit{their}\ 8.90}\right)\left[= \Phi(0.449)\right] = 0.6732\) | M1 | 46, 42 and *their* 4(b) \(\sigma\) substituted in \(\pm\)standardisation formula, condone continuity correction, \(\sigma^2\), \(\sqrt{\sigma}\), and probability found. Condone \(\frac{4}{\textit{their}\ 8.90}\) |
| \(P(\text{both}) = 0.0668 \times 0.6732\) | M1 | Product of *their* 2 probabilities (\(0 <\) both \(< 1\)). Not 0.25 or *their* final answer to 4(a) used |
| \(0.0450\) or \(0.0449\) | A1 | \(0.0449 \leqslant p \leqslant 0.0450\) |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(46 < X < 62) = P\!\left(\frac{46-55}{6} < Z < \frac{62-55}{6}\right)$ | M1 | 46 or 62, 55 and 6 substituted into $\pm$standardisation formula once. Condone $6^2$ and continuity correction $\pm 0.5$ |
| $= P\!\left(-1.5 < Z < \frac{7}{6}\right)$ | B1 | Both standardisation values correct, accept unsimplified |
| $\left[=\Phi\!\left(\frac{7}{6}\right) - (1 - \Phi(1.5))\right] = 0.8784 + (0.9332-1)$ | M1 | Calculating the appropriate area from stated $\Phi$s of $z$-values, must be probabilities |
| $0.812$ | A1 | $0.8115 < p \leqslant 0.812$ |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = \pm 0.674$ | B1 | CAO, critical $z$-value |
| $\frac{36-42}{\sigma} = -0.674$ | M1 | 36 and 42 substituted in $\pm$standardisation formula, no continuity correction, not $\sigma^2$, $\sqrt{\sigma}$, equated to a $z$-value |
| $\sigma = 8.9[0]$ | A1 | WWW. Only dependent on M |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{male} < 46) = 1 - \textit{their}\ 0.9332 = 0.0668$ | M1 | FT value from part **(a)** or Correct: $1 - \Phi\!\left(\frac{46-55}{6}\right)$, condone continuity correction, $\sigma^2$, $\sqrt{\sigma}$, and probability found. Condone unsupported correct value stated |
| $P(\text{female} < 46) = P\!\left(Z < \frac{46-42}{\textit{their}\ 8.90}\right)\left[= \Phi(0.449)\right] = 0.6732$ | M1 | 46, 42 and *their* 4(b) $\sigma$ substituted in $\pm$standardisation formula, condone continuity correction, $\sigma^2$, $\sqrt{\sigma}$, and probability found. Condone $\frac{4}{\textit{their}\ 8.90}$ |
| $P(\text{both}) = 0.0668 \times 0.6732$ | M1 | Product of *their* 2 probabilities ($0 <$ both $< 1$). Not 0.25 or *their* final answer to **4(a)** used |
| $0.0450$ or $0.0449$ | A1 | $0.0449 \leqslant p \leqslant 0.0450$ |4 The weights of male leopards in a particular region are normally distributed with mean 55 kg and standard deviation 6 kg .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen male leopard from this region weighs between 46 and 62 kg .\\
The weights of female leopards in this region are normally distributed with mean 42 kg and standard deviation $\sigma \mathrm { kg }$. It is known that $25 \%$ of female leopards in the region weigh less than 36 kg .
\item Find the value of $\sigma$.\\
The distributions of the weights of male and female leopards are independent of each other. A male leopard and a female leopard are each chosen at random.
\item Find the probability that both the weights of these leopards are less than 46 kg .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q4 [11]}}