| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Justify normal approximation |
| Difficulty | Moderate -0.8 This is a straightforward binomial distribution question with routine normal approximation. Part (i) is direct binomial probability calculation, part (ii) requires stating the np>5 and nq>5 criteria (which fails here), and part (iii) is standard normal approximation with continuity correction. All techniques are textbook exercises requiring only recall and basic application, no problem-solving insight needed. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 2) = (0.25)^2 \times (0.75)^x \times {}^8C_2 = 0.311\) | M1, A1 | 3 term binomial expression involving \({}^8C_?\) something, powers summing to 8; correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(12 \times 0.25 = 3, < 5\) so not possible | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| mean \(= 40 \times 0.25 (= 10)\) and variance \(= 40 \times 0.25 \times 0.75 (= 7.5)\) | B1 | \(40 \times 0.25\) and \(40 \times 0.25 \times 0.75\) seen, o.e. |
| \(P(X \text{ at least } 13) = P\left(z \geq \frac{12.5 - 10}{\sqrt{7.5}}\right)\) | M1 | standardising, \(\pm\), with or without cc, must have sq rt |
| \(= P(z > 0.913) = 1 - \Phi(0.913) = 1 - 0.8194 = 0.181\) | M1, M1, A1 | continuity correction 12.5 or 13.5 correct area, i.e. < 0.5 legit; correct answer |
**(i) Binomial probability:**
$P(X = 2) = (0.25)^2 \times (0.75)^x \times {}^8C_2 = 0.311$ | M1, A1 | 3 term binomial expression involving ${}^8C_?$ something, powers summing to 8; correct answer | **[2]**
**(ii) Feasibility check:**
$12 \times 0.25 = 3, < 5$ so not possible | B1 | | **[1]**
**(iii) Normal approximation:**
mean $= 40 \times 0.25 (= 10)$ and variance $= 40 \times 0.25 \times 0.75 (= 7.5)$ | B1 | $40 \times 0.25$ and $40 \times 0.25 \times 0.75$ seen, o.e. |
$P(X \text{ at least } 13) = P\left(z \geq \frac{12.5 - 10}{\sqrt{7.5}}\right)$ | M1 | standardising, $\pm$, with or without cc, must have sq rt |
$= P(z > 0.913) = 1 - \Phi(0.913) = 1 - 0.8194 = 0.181$ | M1, M1, A1 | continuity correction 12.5 or 13.5 correct area, i.e. < 0.5 legit; correct answer | **[5]**
---5 In the holidays Martin spends $25 \%$ of the day playing computer games. Martin's friend phones him once a day at a randomly chosen time.\\
(i) Find the probability that, in one holiday period of 8 days, there are exactly 2 days on which Martin is playing computer games when his friend phones.\\
(ii) Another holiday period lasts for 12 days. State with a reason whether it is appropriate to use a normal approximation to find the probability that there are fewer than 7 days on which Martin is playing computer games when his friend phones.\\
(iii) Find the probability that there are at least 13 days of a 40-day holiday period on which Martin is playing computer games when his friend phones.
\hfill \mbox{\textit{CAIE S1 2010 Q5 [8]}}