| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.8 This is a straightforward application of standard binomial and normal approximation techniques. Part (a) requires basic binomial probability calculation with n=12, part (b) applies the routine normal approximation with continuity correction for n=80, and part (c) asks for the standard np>5 and nq>5 justification. All steps are textbook procedures with no problem-solving insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([1 - P(10,11,12) =]\) \(1 - \left({}^{12}C_{10}\ 0.9^{10}\ 0.1^2 + {}^{12}C_{11}\ 0.9^{11}\ 0.1^1 + {}^{12}C_{12}\ 0.9^{12}\ 0.1^0\right)\) | M1 | One term \({}^{12}C_x\ p^x(1-p)^{12-x}\), for \(0 < x < 12\), \(0 < p < 1\) |
| \(= 1 - (0.230128 + 0.376573 + 0.282430)\) | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer |
| \(0.111\) | B1 | Mark the final answer at the most accurate value, \(0.1108 < p \leqslant 0.111\) WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(0,1,2,3,4,5,6,7,8,9) =]\) \({}^{12}C_0\ 0.9^0\ 0.1^{12} + {}^{12}C_1\ 0.9^1\ 0.1^{11} + \ldots + {}^{12}C_9\ 0.9^9\ 0.1^3\) | M1 | One term \({}^{12}C_x\ p^x(1-p)^{12-x}\), for \(0 < x < 12\), \(0 < p < 1\) |
| (full sum of 10 terms) | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer. If answer correct condone omission of any 7 of the 8 middle terms |
| \(0.111\) | B1 | Final answer \(0.1108 < p \leqslant 0.111\) WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{Mean} = 80 \times 0.9 =]\ 72\), \([\text{Variance} = 80 \times 0.9 \times 0.1] = 7.2\) | B1 | 72 and 7.2 seen, allow unsimplified. May be seen in standardisation formula. \((2.683 \leqslant \sigma < 2.684\) imply correct variance) |
| \(P(X > 69) = P\!\left(Z > \frac{69.5 - 72}{\sqrt{7.2}}\right)\) | M1 | Substituting *their* mean and \(\sqrt{\textit{their}\ \text{variance}}\) into \(\pm\)standardisation formula (any number for 69.5), not *their* 7.2, not \(\sqrt{\textit{their}}\ 2.683\) |
| M1 | Using continuity correction 69.5 or 68.5 in *their* standardisation formula | |
| \([= P(Z > -0.9317) =]\ \Phi(0.9317)\) | M1 | Appropriate area \(\Phi\), from final process, must be probability |
| \(0.824\) | A1 | \(0.8239 \leqslant p \leqslant 0.8243\) WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(np = 72,\ nq = 8\) Both greater than 5, [so approximation is valid] | B1 | \(np\), \(nq\) evaluated accurately. Both \(np\) & \(nq\) referenced correctly. \(> 5\) or greater than 5 seen |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[1 - P(10,11,12) =]$ $1 - \left({}^{12}C_{10}\ 0.9^{10}\ 0.1^2 + {}^{12}C_{11}\ 0.9^{11}\ 0.1^1 + {}^{12}C_{12}\ 0.9^{12}\ 0.1^0\right)$ | M1 | One term ${}^{12}C_x\ p^x(1-p)^{12-x}$, for $0 < x < 12$, $0 < p < 1$ |
| $= 1 - (0.230128 + 0.376573 + 0.282430)$ | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer |
| $0.111$ | B1 | Mark the final answer at the most accurate value, $0.1108 < p \leqslant 0.111$ WWW |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(0,1,2,3,4,5,6,7,8,9) =]$ ${}^{12}C_0\ 0.9^0\ 0.1^{12} + {}^{12}C_1\ 0.9^1\ 0.1^{11} + \ldots + {}^{12}C_9\ 0.9^9\ 0.1^3$ | M1 | One term ${}^{12}C_x\ p^x(1-p)^{12-x}$, for $0 < x < 12$, $0 < p < 1$ |
| (full sum of 10 terms) | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer. If answer correct condone omission of any 7 of the 8 middle terms |
| $0.111$ | B1 | Final answer $0.1108 < p \leqslant 0.111$ WWW |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{Mean} = 80 \times 0.9 =]\ 72$, $[\text{Variance} = 80 \times 0.9 \times 0.1] = 7.2$ | B1 | 72 and 7.2 seen, allow unsimplified. May be seen in standardisation formula. $(2.683 \leqslant \sigma < 2.684$ imply correct variance) |
| $P(X > 69) = P\!\left(Z > \frac{69.5 - 72}{\sqrt{7.2}}\right)$ | M1 | Substituting *their* mean and $\sqrt{\textit{their}\ \text{variance}}$ into $\pm$standardisation formula (any number for 69.5), not *their* 7.2, not $\sqrt{\textit{their}}\ 2.683$ |
| | M1 | Using continuity correction 69.5 or 68.5 in *their* standardisation formula |
| $[= P(Z > -0.9317) =]\ \Phi(0.9317)$ | M1 | Appropriate area $\Phi$, from final process, must be probability |
| $0.824$ | A1 | $0.8239 \leqslant p \leqslant 0.8243$ WWW |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $np = 72,\ nq = 8$ Both greater than 5, [so approximation is valid] | B1 | $np$, $nq$ evaluated accurately. **Both** $np$ & $nq$ referenced correctly. $> 5$ or greater than 5 seen |
---6 At a company's call centre, $90 \%$ of callers are connected immediately to a representative.\\
A random sample of 12 callers is chosen.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that fewer than 10 of these callers are connected immediately.\\
A random sample of 80 callers is chosen.
\item Use an approximation to find the probability that more than 69 of these callers are connected immediately.
\item Justify the use of your approximation in part (b).
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q6 [9]}}