| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Single probability inequality |
| Difficulty | Moderate -0.8 This is a straightforward application of normal approximation to binomial with continuity correction. Part (a) requires identifying n=110, p=0.25, applying continuity correction (X<22 becomes X≤21.5), standardizing to find a z-score, and reading from tables. The setup is clear, the method is standard, and it's a routine S1 topic requiring no problem-solving insight—just procedural execution of a well-practiced technique. |
| Spec | 2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= 110\times0.25 = 27.5\); Variance \(= 110\times0.25\times0.75 = 20.625 = \frac{165}{8}\) | B1 | 27.5 and 20.625 seen, allow unsimplified. \(\sqrt{\frac{165}{8}}\) or \(\frac{\sqrt{330}}{4}\) (4.5414... to at least 4sf) implies correct variance. Penalise incorrect identification, condone no identification. |
| \(P(X<22) = P\!\left(Z < \frac{21.5-27.5}{\sqrt{20.625}}\right)\) | M1 | Substituting their 27.5 and their 20.625 into \(\pm\)standardising formula (any number for 21.5), not \(\sigma^2\), not \(\sqrt{\sigma}\). |
| M1 | Using continuity correction 21.5 or 22.5 in their standardisation formula. | |
| \([P(Z < -1.3212)] = 1-\Phi(1.3212)\) \(= 1-0.9068\) | M1 | Appropriate probability area from final process, must be a probability. May be implied by sketch. Expect final answer \(< 0.5\). |
| \(0.0932\) | A1 | \(0.0932 \leq p < 0.09325\). If either M1 M1 not awarded, SC B1 \(0.0932 \leq p < 0.09325\) WWW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Method 1: \([1-P(8,9,10)]\) \(= 1-({}^{10}C_8\,0.85^8\,0.15^2+{}^{10}C_9\,0.85^9\,0.15^1+0.85^{10})\) \(=[1-(0.275897+0.347425+0.196874)]\) | M1 | One term \({}^{10}C_x\,(p)^x(1-p)^{10-x}\), \(0 |
| A1 | Correct unsimplified expression. Condone omission of last bracket only. | |
| \(= 0.180\) | B1 | \(0.1795 < p \leq 0.180\) |
| Method 2: \([P(0,1,2,3,4,5,6,7)]\) \(= 0.15^{10}+{}^{10}C_1\,0.85\times0.15^9+\ldots+{}^{10}C_7\,0.85^7\,0.15^3\) | (M1) | One term \({}^{10}C_x\,(p)^x(1-p)^{10-x}\), \(0 |
| (A1) | Correct unsimplified expression. | |
| \(= 0.180\) | (B1) | \(0.1795 < p \leq 0.180\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.25\times0.6\times0.15\times6\) | M1 | \(0.25\times0.6\times0.15\times k\), \(k\) an integer \(> 1\). |
| \(0.135,\ \frac{27}{200}\) | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 110\times0.25 = 27.5$; Variance $= 110\times0.25\times0.75 = 20.625 = \frac{165}{8}$ | B1 | 27.5 and 20.625 seen, allow unsimplified. $\sqrt{\frac{165}{8}}$ or $\frac{\sqrt{330}}{4}$ (4.5414... to at least 4sf) implies correct variance. Penalise incorrect identification, condone no identification. |
| $P(X<22) = P\!\left(Z < \frac{21.5-27.5}{\sqrt{20.625}}\right)$ | M1 | Substituting their 27.5 and their 20.625 into $\pm$standardising formula (any number for 21.5), not $\sigma^2$, not $\sqrt{\sigma}$. |
| | M1 | Using continuity correction 21.5 or 22.5 in their standardisation formula. |
| $[P(Z < -1.3212)] = 1-\Phi(1.3212)$ $= 1-0.9068$ | M1 | Appropriate probability area from final process, must be a probability. May be implied by sketch. Expect final answer $< 0.5$. |
| $0.0932$ | A1 | $0.0932 \leq p < 0.09325$. If either M1 M1 not awarded, SC B1 $0.0932 \leq p < 0.09325$ WWW. |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Method 1:** $[1-P(8,9,10)]$ $= 1-({}^{10}C_8\,0.85^8\,0.15^2+{}^{10}C_9\,0.85^9\,0.15^1+0.85^{10})$ $=[1-(0.275897+0.347425+0.196874)]$ | M1 | One term ${}^{10}C_x\,(p)^x(1-p)^{10-x}$, $0<p<1$, $x\neq0$ or 10. |
| | A1 | Correct unsimplified expression. Condone omission of last bracket only. |
| $= 0.180$ | B1 | $0.1795 < p \leq 0.180$ |
| **Method 2:** $[P(0,1,2,3,4,5,6,7)]$ $= 0.15^{10}+{}^{10}C_1\,0.85\times0.15^9+\ldots+{}^{10}C_7\,0.85^7\,0.15^3$ | (M1) | One term ${}^{10}C_x\,(p)^x(1-p)^{10-x}$, $0<p<1$, $x\neq0$ or 10. |
| | (A1) | Correct unsimplified expression. |
| $= 0.180$ | (B1) | $0.1795 < p \leq 0.180$ |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.25\times0.6\times0.15\times6$ | M1 | $0.25\times0.6\times0.15\times k$, $k$ an integer $> 1$. |
| $0.135,\ \frac{27}{200}$ | A1 | |
---6 The residents of Mahjing were asked to classify their local bus service:
\begin{itemize}
\item $25 \%$ of residents classified their service as good.
\item $60 \%$ of residents classified their service as satisfactory.
\item $15 \%$ of residents classified their service as poor.
\begin{enumerate}[label=(\alph*)]
\item A random sample of 110 residents of Mahjing is chosen.
\end{itemize}
Use a suitable approximation to find the probability that fewer than 22 residents classified their bus service as good.
\item For a random sample of 10 residents of Mahjing, find the probability that fewer than 8 classified their bus service as good or satisfactory.
\item Three residents of Mahjing are selected at random.
Find the probability that one resident classified the bus service as good, one as satisfactory and one as poor.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q6 [10]}}