| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| 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 | Standard +0.3 This is a straightforward application of binomial distribution (part i) and normal approximation to binomial (part ii). Part (i) requires calculating P(X > 2) for B(14, 0.09) using complement rule. Part (ii) is a standard normal approximation with continuity correction for B(200, 0.76). Both are routine textbook exercises requiring no novel insight, though slightly easier than average due to clear setup and standard techniques. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - P(0,1,2)\) | M1 | For \(1 - P(0,1,2)\) |
| \(= 1-\big((0.91)^{14} + (0.09)(0.91)^{13}\times{}_{14}C_1 + (0.09)^2(0.91)^{12}\times{}_{14}C_2\big)\) | B1 | Correct numerical expression for \(P(0)\) or \(P(1)\) |
| B1 | Correct numerical expression for \(P(2)\) | |
| \(= 1-(0.2670+0.3698+0.2377)\) | ||
| \(= 0.126\) | A1 | 4 Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mu = 200\times 0.76 = 152\), \(\sigma^2 = 200\times 0.76\times 0.24 = 36.48\) | B1 | For both mean and variance correct |
| \(P(X>155)\) | M1 | For standardising, with or without cc; must have \(\sqrt{\phantom{x}}\) on denom |
| \(= 1-\Phi\!\left(\dfrac{155.5-152}{\sqrt{36.48}}\right) = 1-\Phi(1.5795)\) | M1 | For use of continuity correction \(154.5\) or \(155.5\) |
| M1 | For finding an area \(< 0.5\) from their \(z\) | |
| \(= 1-0.7188 = 0.281\) | A1 | 5 For answer rounding to \(0.281\) |
## Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - P(0,1,2)$ | M1 | For $1 - P(0,1,2)$ |
| $= 1-\big((0.91)^{14} + (0.09)(0.91)^{13}\times{}_{14}C_1 + (0.09)^2(0.91)^{12}\times{}_{14}C_2\big)$ | B1 | Correct numerical expression for $P(0)$ or $P(1)$ |
| | B1 | Correct numerical expression for $P(2)$ |
| $= 1-(0.2670+0.3698+0.2377)$ | | |
| $= 0.126$ | A1 | **4** Correct answer |
## Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mu = 200\times 0.76 = 152$, $\sigma^2 = 200\times 0.76\times 0.24 = 36.48$ | B1 | For both mean and variance correct |
| $P(X>155)$ | M1 | For standardising, with or without cc; must have $\sqrt{\phantom{x}}$ on denom |
| $= 1-\Phi\!\left(\dfrac{155.5-152}{\sqrt{36.48}}\right) = 1-\Phi(1.5795)$ | M1 | For use of continuity correction $154.5$ or $155.5$ |
| | M1 | For finding an area $< 0.5$ from their $z$ |
| $= 1-0.7188 = 0.281$ | A1 | **5** For answer rounding to $0.281$ |7 A survey of adults in a certain large town found that $76 \%$ of people wore a watch on their left wrist, $15 \%$ wore a watch on their right wrist and $9 \%$ did not wear a watch.\\
(i) A random sample of 14 adults was taken. Find the probability that more than 2 adults did not wear a watch.\\
(ii) A random sample of 200 adults was taken. Using a suitable approximation, find the probability that more than 155 wore a watch on their left wrist.
\hfill \mbox{\textit{CAIE S1 2006 Q7 [9]}}