| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Standard +0.3 This is a straightforward application of binomial probability (part a) and normal approximation to binomial (part b). Part (a) requires calculating P(3≤X≤7) for B(8, 0.48), which is routine but involves multiple probability calculations. Part (b) is a standard normal approximation with continuity correction for B(125, 0.52). Both parts follow textbook procedures with no novel insight required, making this slightly easier than average for an S1 question. |
| Spec | 2.04d Normal approximation to binomial5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(3,4,\ldots7) = 1 - P(0,1,2,8)]\) \(= 1-({}^8C_0\ 0.48^0\ 0.52^8 + {}^8C_1\ 0.48^1\ 0.52^7 + {}^8C_2\ 0.48^2\ 0.52^6 + {}^8C_8\ 0.48^8\ 0.52^0)\) | M1 | One term \({}^8C_x\ p^x(1-p)^{8-x}\), for \(0 < x < 8\), \(0 < p < 1\) |
| \(= 1-(0.00534597 + 0.039478 + 0.127544 + 0.0028179)\) | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer |
| \(0.825\) | B1 | Mark the final answer at the most accurate value. \(0.8248 < p \leqslant 0.825\) WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(3,4,5,6,7) =]\) \({}^8C_3\ 0.48^3\ 0.52^5 + {}^8C_4\ 0.48^4\ 0.52^4 + {}^8C_5\ 0.48^5\ 0.52^3 + {}^8C_6\ 0.48^6\ 0.52^2 + {}^8C_7\ 0.48^7\ 0.52^1\) | M1 | One term \({}^8C_x\ p^x(1-p)^{8-x}\), for \(0 < x < 8\), \(0 < p < 1\) |
| A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer | |
| \(0.825\) | B1 | Final answer \(0.8248 < p \leqslant 0.825\) WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{Mean} = 0.52 \times 125 =] 65\), \([\text{var} = 0.52 \times 0.48 \times 125 =] 31.2\) | B1 | 65 and 31.2 seen, allow unsimplified. May be seen in standardisation formula. (\(5.585 < \sigma \leqslant 5.586\) imply correct variance) |
| \([P(X>72) =]\ P\!\left(Z > \dfrac{72.5-65}{\sqrt{31.2}}\right)\ [= P(Z > 1.343)]\) | M1 | Substituting *their* 65 and \(\sqrt{\textit{their}\ 31.2}\) into \(\pm\)standardisation formula (any number for 72·5), not *their* 31.2, \(\sqrt{\textit{their}\ 5.586}\) |
| M1 | Using continuity correction 72·5 or 71·5 in *their* standardisation formula. Note \(\dfrac{\pm7.5}{\sqrt{31.2}}\) or \(\dfrac{\pm7.5}{5.586}\) seen gains M2 BOD | |
| \(= 1 - 0.9104\) | M1 | Appropriate area \(\Phi\), from final process, must be probability |
| \(0.0896\) | A1 | \(0.0896 \leqslant p \leqslant 0.0897\) WWW |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(3,4,\ldots7) = 1 - P(0,1,2,8)]$ $= 1-({}^8C_0\ 0.48^0\ 0.52^8 + {}^8C_1\ 0.48^1\ 0.52^7 + {}^8C_2\ 0.48^2\ 0.52^6 + {}^8C_8\ 0.48^8\ 0.52^0)$ | M1 | One term ${}^8C_x\ p^x(1-p)^{8-x}$, for $0 < x < 8$, $0 < p < 1$ |
| $= 1-(0.00534597 + 0.039478 + 0.127544 + 0.0028179)$ | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer |
| $0.825$ | B1 | Mark the final answer at the most accurate value. $0.8248 < p \leqslant 0.825$ WWW |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(3,4,5,6,7) =]$ ${}^8C_3\ 0.48^3\ 0.52^5 + {}^8C_4\ 0.48^4\ 0.52^4 + {}^8C_5\ 0.48^5\ 0.52^3 + {}^8C_6\ 0.48^6\ 0.52^2 + {}^8C_7\ 0.48^7\ 0.52^1$ | M1 | One term ${}^8C_x\ p^x(1-p)^{8-x}$, for $0 < x < 8$, $0 < p < 1$ |
| | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer |
| $0.825$ | B1 | Final answer $0.8248 < p \leqslant 0.825$ WWW |
---
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{Mean} = 0.52 \times 125 =] 65$, $[\text{var} = 0.52 \times 0.48 \times 125 =] 31.2$ | B1 | 65 and 31.2 seen, allow unsimplified. May be seen in standardisation formula. ($5.585 < \sigma \leqslant 5.586$ imply correct variance) |
| $[P(X>72) =]\ P\!\left(Z > \dfrac{72.5-65}{\sqrt{31.2}}\right)\ [= P(Z > 1.343)]$ | M1 | Substituting *their* 65 and $\sqrt{\textit{their}\ 31.2}$ into $\pm$standardisation formula (any number for 72·5), not *their* 31.2, $\sqrt{\textit{their}\ 5.586}$ |
| | M1 | Using continuity correction 72·5 or 71·5 in *their* standardisation formula. Note $\dfrac{\pm7.5}{\sqrt{31.2}}$ or $\dfrac{\pm7.5}{5.586}$ seen gains M2 BOD |
| $= 1 - 0.9104$ | M1 | Appropriate area $\Phi$, from final process, must be probability |
| $0.0896$ | A1 | $0.0896 \leqslant p \leqslant 0.0897$ WWW |
---2 The residents of Persham were surveyed about the reliability of their internet service. 12\% rated the service as 'poor', $36 \%$ rated it as 'satisfactory' and $52 \%$ rated it as 'good'.
A random sample of 8 residents of Persham is chosen.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that more than 2 and fewer than 8 of them rate their internet service as poor or satisfactory.\\
A random sample of 125 residents of Persham is now chosen.
\item Use an approximation to find the probability that more than 72 of these residents rate their internet service as good.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q2 [8]}}