| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | March |
| 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-to-normal approximation (part c). All parts are routine applications of textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X > 1.26) = P\!\left(Z > \dfrac{1.26 - 1.20}{0.16}\right)]\) \([= P(Z > 0.375)]\) | M1 | Use of \(\pm\) standardisation formula with 1.26, 1.20 and 0.16, not \(\sigma^2\), not \(\sigma\), no continuity correction |
| \(= 1 - 0.6462\) | M1 | Calculating the appropriate probability area (leading to final probability, expect \(< 0.5\)) |
| \(= 0.354\) | A1 | \(0.3538,\ 0.3535 < p \leqslant 0.354\). Only dependent on 2nd M mark. SC B1 for \(0.3535 < p \leqslant 0.354\) if M0M0A0 awarded |
| \([500 \times \textit{their}\ 0.3538] = 176, 177\) | B1 FT | Strict FT *their* at least 4-figure calculated probability, seen anywhere (not a \(z\)-value). Final answer must be a single positive integer value, no approximation or rounding stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[P\!\left(Z > \dfrac{2.40 - 2.50}{\sigma}\right) = 0.20\right]\) | B1 | \(-0.842 \leqslant z < -0.8415\) or \(0.8415 < z \leqslant 0.842\) seen |
| \(\dfrac{2.40 - 2.50}{\sigma} = -0.842\) | M1 | Use of \(\pm\) standardisation formula with 2.40, 2.50, \(\sigma\) and a \(z\)-value (not 0.20, 0.80, 0.158, 0.7881, 0.2119, 0.5793, 0.4207, …), not \(\sigma^2\). Condone continuity correction of \(\pm0.005\). Condone \(-\dfrac{0.1}{\sigma} = -0.842\) etc. for M1 |
| \(\sigma = 0.119\) | A1 | \(0.1185 < \sigma \leqslant 0.119\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Mean} = 80 \times 0.2 = 16\) | B1 | 16 and 12.8 seen, allow unsimplified. May be seen in standardisation formula. \(\frac{8\sqrt{5}}{5}\), 3.5777… to at least three significant figures implies correct variance. Incorrect notation penalised. |
| \(\text{Variance} = 80 \times 0.2 \times 0.8 = 12.8\) | ||
| \(P(X < 22) = P\left(Z < \frac{21.5 - 16}{\sqrt{12.8}}\right)\) | M1 | Substituting *their* 16 (not 1.2, 2.5) and *their* 12.8 (not 0.16, *their* 0.119) in the \(\pm\)standardising formula (any number for 21.5), condone \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \([P(Z < 1.537) = \Phi(1.537)]\) | M1 | Using continuity correction 21.5 or 22.5 in *their* standardisation formula. |
| M1 | Appropriate area \(\Phi\), from final process, must be a probability. | |
| 0.938 | A1 | \(0.9375 < p \leqslant 0.938\) |
| Total | 5 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X > 1.26) = P\!\left(Z > \dfrac{1.26 - 1.20}{0.16}\right)]$ $[= P(Z > 0.375)]$ | M1 | Use of $\pm$ standardisation formula with 1.26, 1.20 and 0.16, not $\sigma^2$, not $\sigma$, no continuity correction |
| $= 1 - 0.6462$ | M1 | Calculating the appropriate probability area (leading to final probability, expect $< 0.5$) |
| $= 0.354$ | A1 | $0.3538,\ 0.3535 < p \leqslant 0.354$. Only dependent on 2nd M mark. SC B1 for $0.3535 < p \leqslant 0.354$ if M0M0A0 awarded |
| $[500 \times \textit{their}\ 0.3538] = 176, 177$ | B1 FT | Strict FT *their* at least 4-figure calculated probability, seen anywhere (not a $z$-value). Final answer must be a single positive integer value, no approximation or rounding stated |
**Total: 4 marks**
---
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[P\!\left(Z > \dfrac{2.40 - 2.50}{\sigma}\right) = 0.20\right]$ | B1 | $-0.842 \leqslant z < -0.8415$ or $0.8415 < z \leqslant 0.842$ seen |
| $\dfrac{2.40 - 2.50}{\sigma} = -0.842$ | M1 | Use of $\pm$ standardisation formula with 2.40, 2.50, $\sigma$ and a $z$-value (not 0.20, 0.80, 0.158, 0.7881, 0.2119, 0.5793, 0.4207, …), not $\sigma^2$. Condone continuity correction of $\pm0.005$. Condone $-\dfrac{0.1}{\sigma} = -0.842$ etc. for M1 |
| $\sigma = 0.119$ | A1 | $0.1185 < \sigma \leqslant 0.119$ |
**Total: 3 marks**
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Mean} = 80 \times 0.2 = 16$ | B1 | 16 and 12.8 seen, allow unsimplified. May be seen in standardisation formula. $\frac{8\sqrt{5}}{5}$, 3.5777… to at least three significant figures implies correct variance. Incorrect notation penalised. |
| $\text{Variance} = 80 \times 0.2 \times 0.8 = 12.8$ | | |
| $P(X < 22) = P\left(Z < \frac{21.5 - 16}{\sqrt{12.8}}\right)$ | M1 | Substituting *their* 16 (not 1.2, 2.5) and *their* 12.8 (not 0.16, *their* 0.119) in the $\pm$standardising formula (any number for 21.5), condone $\sigma^2$ or $\sqrt{\sigma}$ |
| $[P(Z < 1.537) = \Phi(1.537)]$ | M1 | Using continuity correction 21.5 or 22.5 in *their* standardisation formula. |
| | M1 | Appropriate area $\Phi$, from final process, must be a probability. |
| 0.938 | A1 | $0.9375 < p \leqslant 0.938$ |
| **Total** | **5** | |
---4 A company sells small and large bags of rice. The masses of the small bags of rice are normally distributed with mean 1.20 kg and standard deviation 0.16 kg .
\begin{enumerate}[label=(\alph*)]
\item In a random sample of 500 of these small bags of rice, how many would you expect to have a mass greater than 1.26 kg ?\\
The masses of the large bags of rice are normally distributed with mean 2.50 kg and standard deviation $\sigma \mathrm { kg } .20 \%$ of these large bags of rice have a mass less than 2.40 kg .
\item Find the value of $\sigma$.\\
A random sample of 80 large bags of rice is chosen.
\item Use a suitable approximation to find the probability that fewer than 22 of these large bags of rice have a mass less than 2.40 kg .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q4 [12]}}