| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Normal approximation to binomial |
| Difficulty | Standard +0.3 Part (i) is a straightforward binomial probability calculation with small n requiring P(X≤2). Part (ii) is a standard normal approximation to binomial with continuity correction—routine application of a well-practiced technique. Both parts are textbook exercises with no problem-solving insight required, making this slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0,1,2) = (0.66)^{14} + {}^{14}C_1(0.34)(0.66)^{13} + {}^{14}C_2(0.34)^2(0.66)^{12}\) | M1 | Binomial term of form \({}^{14}C_x p^x (1-p)^{14-x}\), \(0 < p < 1\) any \(p\), \(x \neq 14, 0\) |
| \(= 0.0029758 + 0.02146239 + 0.071866\) | A1 | Correct unsimplified answer |
| \(= 0.0963\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 600 \times 0.34 = 204\), Var \(= 600 \times 0.34 \times 0.66 = 134.64\) | B1 | Correct unsimplified \(np\) and \(npq\) (or sd \(= 11.603\) or Variance \(= 3366/25\)) |
| \(P(<190) = P\!\left(z < \dfrac{189.5 - 204}{\sqrt{134.64}}\right) = P(z < -1.2496)\) | M1 | Substituting their \(\mu\) and \(\sigma\) (no \(\sigma^2\) or \(\sqrt{\sigma}\)) into Standardisation Formula with a numerical value for '\(189.5\)'. Condone \(\pm\) standardisation formula |
| M1 | Using continuity correction 189.5 or 190.5 within a Standardisation formula | |
| \(= 1 - \Phi(1.2496)\) | M1 | Appropriate area \(\Phi\) from standardisation formula \(P(z<\ldots)\) in final solution, (\(<0.5\) if \(z\) is \(-\)ve, \(>0.5\) if \(z\) is \(+\)ve) |
| \(= 1 - 0.8944 = 0.106\) | A1 | Correct final answer |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0,1,2) = (0.66)^{14} + {}^{14}C_1(0.34)(0.66)^{13} + {}^{14}C_2(0.34)^2(0.66)^{12}$ | M1 | Binomial term of form ${}^{14}C_x p^x (1-p)^{14-x}$, $0 < p < 1$ any $p$, $x \neq 14, 0$ |
| $= 0.0029758 + 0.02146239 + 0.071866$ | A1 | Correct unsimplified answer |
| $= 0.0963$ | A1 | Correct answer |
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## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 600 \times 0.34 = 204$, Var $= 600 \times 0.34 \times 0.66 = 134.64$ | B1 | Correct unsimplified $np$ and $npq$ (or sd $= 11.603$ or Variance $= 3366/25$) |
| $P(<190) = P\!\left(z < \dfrac{189.5 - 204}{\sqrt{134.64}}\right) = P(z < -1.2496)$ | M1 | Substituting their $\mu$ and $\sigma$ (no $\sigma^2$ or $\sqrt{\sigma}$) into Standardisation Formula with a numerical value for '$189.5$'. Condone $\pm$ standardisation formula |
| | M1 | Using continuity correction 189.5 or 190.5 within a Standardisation formula |
| $= 1 - \Phi(1.2496)$ | M1 | Appropriate area $\Phi$ from standardisation formula $P(z<\ldots)$ in final solution, ($<0.5$ if $z$ is $-$ve, $>0.5$ if $z$ is $+$ve) |
| $= 1 - 0.8944 = 0.106$ | A1 | Correct final answer |5 On average, $34 \%$ of the people who go to a particular theatre are men.\\
(i) A random sample of 14 people who go to the theatre is chosen. Find the probability that at most 2 people are men.\\
(ii) Use an approximation to find the probability that, in a random sample of 600 people who go to the theatre, fewer than 190 are men.\\
\hfill \mbox{\textit{CAIE S1 2019 Q5 [8]}}