| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Moderate -0.3 This is a standard two-stage normal distribution problem: (a) uses two given percentiles to find μ and σ by solving simultaneous equations with z-scores (routine S1 technique), (b) applies binomial distribution with probability from part (a). Both parts are textbook exercises requiring no novel insight, though the calculation is slightly more involved than single-percentile problems. |
| Spec | 2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{182.7 - \mu}{\sigma} = 1.282\) | B1 | 1.282 or \(-1.282\) seen, CAO (critical value). |
| \(\frac{162.5 - \mu}{\sigma} = -0.253\) | B1 | \(-0.2535 < z \leqslant -0.253\) or \(0.253 \leqslant z < 0.2535\) seen. |
| M1 | One standardisation formula, not \(\sigma^2\), or \(\sqrt{\sigma}\), with 182.7 or 162.5 substituted correctly equated to a \(z\) value (not 0.9, 0.1, 0.8159, 0.5398, 0.4, 0.6, 0.6554, 0.7257, …). | |
| Solve, obtaining values for \(\mu\) and \(\sigma\) | M1 | Either a single expression with one variable eliminated formed or two expressions with both variables on the same side seen with at least one variable value stated. |
| \(\mu = 165.8,\ \sigma = 13.2\) | A1 | Answers must be to at least 1 DP (context). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X<8) = 1 - P(8,9,10) =]\ 1 - \left({}^{10}C_8(0.6)^8(0.4)^2 + {}^{10}C_9(0.6)^9(0.4)^1 + (0.6)^{10}\right)\) | M1 | One term \({}^{10}C_x(p)^x(1-p)^{10-x}\). With \(0 < p < 1\), \(x \neq 0\) or 10. |
| \([= 1 - (0.12093 + 0.040311 + 0.0060466)]\) | A1 | Correct unsimplified expression. Allow 10 for \({}^{10}C_9\). Condone omission of last bracket only. If both brackets omitted in unsimplified expression allow recovery for final stated calculation of \(1 - 0.1673\) or final answer WRT 0.8327. |
| \(= 0.833\) | B1 | \(0.8327 < p \leqslant 0.833\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((0.4)^{10} + {}^{10}C_1(0.6)^1(0.4)^9 + {}^{10}C_2(0.6)^2(0.4)^8 + \cdots + {}^{10}C_7(0.6)^7(0.4)^3\) | M1 | One term \({}^{10}C_x(p)^x(1-p)^{10-x}\). With \(0 < p < 1\), \(x \neq 0\) or 10. |
| A1 | Correct unsimplified expression. | |
| \(= 0.833\) | B1 | \(0.8327 < p \leqslant 0.833\). |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{182.7 - \mu}{\sigma} = 1.282$ | B1 | 1.282 or $-1.282$ seen, CAO (critical value). |
| $\frac{162.5 - \mu}{\sigma} = -0.253$ | B1 | $-0.2535 < z \leqslant -0.253$ or $0.253 \leqslant z < 0.2535$ seen. |
| | M1 | One standardisation formula, not $\sigma^2$, or $\sqrt{\sigma}$, with 182.7 or 162.5 substituted correctly equated to a $z$ value (not 0.9, 0.1, 0.8159, 0.5398, 0.4, 0.6, 0.6554, 0.7257, …). |
| Solve, obtaining values for $\mu$ and $\sigma$ | M1 | Either a single expression with one variable eliminated formed or two expressions with both variables on the same side seen with at least one variable value stated. |
| $\mu = 165.8,\ \sigma = 13.2$ | A1 | Answers must be to at least 1 DP (context). |
---
## Question 6(b):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X<8) = 1 - P(8,9,10) =]\ 1 - \left({}^{10}C_8(0.6)^8(0.4)^2 + {}^{10}C_9(0.6)^9(0.4)^1 + (0.6)^{10}\right)$ | M1 | One term ${}^{10}C_x(p)^x(1-p)^{10-x}$. With $0 < p < 1$, $x \neq 0$ or 10. |
| $[= 1 - (0.12093 + 0.040311 + 0.0060466)]$ | A1 | Correct unsimplified expression. Allow 10 for ${}^{10}C_9$. Condone omission of last bracket only. If both brackets omitted in unsimplified expression allow recovery for final stated calculation of $1 - 0.1673$ or final answer WRT 0.8327. |
| $= 0.833$ | B1 | $0.8327 < p \leqslant 0.833$. |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(0.4)^{10} + {}^{10}C_1(0.6)^1(0.4)^9 + {}^{10}C_2(0.6)^2(0.4)^8 + \cdots + {}^{10}C_7(0.6)^7(0.4)^3$ | M1 | One term ${}^{10}C_x(p)^x(1-p)^{10-x}$. With $0 < p < 1$, $x \neq 0$ or 10. |
| | A1 | Correct unsimplified expression. |
| $= 0.833$ | B1 | $0.8327 < p \leqslant 0.833$. |
---6 The heights of the female students at Breven college are normally distributed:
\begin{itemize}
\item $90 \%$ of the female students have heights less than 182.7 cm .
\item $40 \%$ of the female students have heights less than 162.5 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the mean and the standard deviation of the heights of the female students at Breven college.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad3a6a8a-23fe-415a-b2f4-7c49136ccc6c-10_2715_41_110_2008}\\
\includegraphics[max width=\textwidth, alt={}, center]{ad3a6a8a-23fe-415a-b2f4-7c49136ccc6c-11_2723_35_101_20}
\end{itemize}
Ten female students are chosen at random from those at Breven college.
\item Find the probability that fewer than 8 of these 10 students have heights more than 162.5 cm .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q6 [8]}}