| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution parameters found then approximation applied |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: z-score calculation and expected frequency (part a), inverse normal to find unknown parameter (part b), and binomial approximation by normal (part c). All are routine S1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(15.4 < X < 16.8)] = P\!\left(\frac{15.4-16.5}{0.6} < Z < \frac{16.8-16.5}{0.6}\right) = P(-1.833 < Z < 0.5)\) | M1 | Use of \(\pm\) standardisation formula once with 16.5, 0.6 and either 15.4 or 16.8 substituted |
| \(= \Phi(0.5) + \Phi(1.833) - 1\) \(= 0.6915 + 0.9666 - 1\) | M1 | Calculating appropriate probability area (leading to final answer, expect \(> 0.5\)). \(0.6915-(1-0.9666)\) or \((0.6915-0.5)+(0.9666-0.5)\) are alternatives |
| \(= 0.658\) | A1 | \(0.658 \leq p < 0.6585\). If A0 scored, SC B1 for \(0.658 \leqslant p < 0.6585\) |
| Expected number \(= 0.6581 \times 150 = 98, 99\) | B1 FT | FT *their* 4SF (or better) probability. Must be positive single integer. No approximation notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\!\left(Z > \frac{17.1-18.4}{\sigma}\right) = 0.72\) | B1 | \(0.5825 < z \leqslant 0.583\) or \(-0.583 \leqslant z < -0.5825\) seen |
| \(\frac{17.1-18.4}{\sigma} = -0.583\) | M1 | Use of \(\pm\) standardisation formula with 17.1, 18.4, \(\sigma\) and a \(z\)-value (not 0.28, 0.72, 0.4175, 0.2358, 0.7642, 0.6103, 0.3897,…). Condone continuity correct \(\pm 0.05\), not \(\sigma^2\), \(\sqrt{\sigma}\) |
| \(\sigma = 2.23\) | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 120 \times 0.72 = 86.4\); Var \(= 120 \times 0.72 \times 0.28 = 24.192\) | B1 | 86.4, \(84\frac{2}{5}\) and \(24\frac{24}{125}\), 24.192 to at least 3SF seen. May be seen in standardisation formula. \((4.918 \leqslant \sigma \leqslant 4.919\) implies correct variance). Incorrect notation penalised |
| \(P(X < 80) = P\!\left(Z < \frac{79.5-86.4}{\sqrt{24.192}}\right)\) | M1 | Substituting *their* mean (not 18.4) and *their positive* 4.9185 into \(\pm\) standardisation formula (any number for 79.5), condone *their* \(4.918^2\) and \(\sqrt{\textit{their}}\, 4.918\) |
| M1 | Using continuity correction 79.5 or 80.5 in *their* standardisation formula | |
| \([P(Z < -1.4029)] = 1 - \Phi(1.403)\) \(= 1 - 0.9196\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability. Expect final answer \(< 0.5\). Note: correct final answer implies this M1 |
| 0.0804 | A1 | \(0.0803 \leqslant p \leqslant 0.0804\) |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(15.4 < X < 16.8)] = P\!\left(\frac{15.4-16.5}{0.6} < Z < \frac{16.8-16.5}{0.6}\right) = P(-1.833 < Z < 0.5)$ | M1 | Use of $\pm$ standardisation formula once with 16.5, 0.6 and either 15.4 or 16.8 substituted |
| $= \Phi(0.5) + \Phi(1.833) - 1$ $= 0.6915 + 0.9666 - 1$ | M1 | Calculating appropriate probability area (leading to final answer, expect $> 0.5$). $0.6915-(1-0.9666)$ or $(0.6915-0.5)+(0.9666-0.5)$ are alternatives |
| $= 0.658$ | A1 | $0.658 \leq p < 0.6585$. If A0 scored, **SC B1** for $0.658 \leqslant p < 0.6585$ |
| Expected number $= 0.6581 \times 150 = 98, 99$ | B1 FT | FT *their* 4SF (or better) probability. Must be positive single integer. No approximation notation |
**Total: 4 marks**
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\!\left(Z > \frac{17.1-18.4}{\sigma}\right) = 0.72$ | B1 | $0.5825 < z \leqslant 0.583$ or $-0.583 \leqslant z < -0.5825$ seen |
| $\frac{17.1-18.4}{\sigma} = -0.583$ | M1 | Use of $\pm$ standardisation formula with 17.1, 18.4, $\sigma$ and a $z$-value (not 0.28, 0.72, 0.4175, 0.2358, 0.7642, 0.6103, 0.3897,…). Condone continuity correct $\pm 0.05$, not $\sigma^2$, $\sqrt{\sigma}$ |
| $\sigma = 2.23$ | A1 | AWRT |
**Total: 3 marks**
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 120 \times 0.72 = 86.4$; Var $= 120 \times 0.72 \times 0.28 = 24.192$ | B1 | 86.4, $84\frac{2}{5}$ and $24\frac{24}{125}$, 24.192 to at least 3SF seen. May be seen in standardisation formula. $(4.918 \leqslant \sigma \leqslant 4.919$ implies correct variance). Incorrect notation penalised |
| $P(X < 80) = P\!\left(Z < \frac{79.5-86.4}{\sqrt{24.192}}\right)$ | M1 | Substituting *their* mean (not 18.4) and *their positive* 4.9185 into $\pm$ standardisation formula (any number for 79.5), condone *their* $4.918^2$ and $\sqrt{\textit{their}}\, 4.918$ |
| | M1 | Using continuity correction 79.5 or 80.5 in *their* standardisation formula |
| $[P(Z < -1.4029)] = 1 - \Phi(1.403)$ $= 1 - 0.9196$ | M1 | Appropriate area $\Phi$, from final process, must be a probability. Expect final answer $< 0.5$. Note: correct final answer implies this M1 |
| 0.0804 | A1 | $0.0803 \leqslant p \leqslant 0.0804$ |
**Total: 5 marks**
---5 The lengths of Western bluebirds are normally distributed with mean 16.5 cm and standard deviation 0.6 cm .
A random sample of 150 of these birds is selected.
\begin{enumerate}[label=(\alph*)]
\item How many of these 150 birds would you expect to have length between 15.4 cm and 16.8 cm ?\\
The lengths of Eastern bluebirds are normally distributed with mean 18.4 cm and standard deviation $\sigma \mathrm { cm }$. It is known that $72 \%$ of Eastern bluebirds have length greater than 17.1 cm .
\item Find the value of $\sigma$.\\
A random sample of 120 Eastern bluebirds is chosen.
\item Use an approximation to find the probability that fewer than 80 of these 120 bluebirds have length greater than 17.1 cm .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q5 [12]}}