| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Probability between two values |
| Difficulty | Standard +0.3 Part (i) is a standard normal approximation to binomial with continuity correction—routine for S1 level. Part (ii) requires Bayes' theorem with three events, which is slightly more demanding than basic conditional probability but still a well-practiced technique. Overall slightly easier than average A-level due to straightforward setup and standard methods. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = 0.2\); \(\mu = 96 \times 0.2 = 19.2\); \(\sigma^2 = 96 \times 0.2 \times 0.8 = 15.36\) | B1 | \(96 \times 0.2\) and \(96 \times 0.2 \times 0.8\) seen |
| \(P(<20) = P\!\left(z < \frac{19.5 - 19.2}{\sqrt{15.36}}\right) = P(z < 0.07654)\) | M1 | Standardising must have sq rt |
| M1 | Continuity correction either 19.5 or 20.5 | |
| M1 | Correct area \((> 0.5)\) | |
| \(= 0.531\) | A1 [5] | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(OT | B) = \frac{0.2 \times 0.6}{0.05 \times 0.3 + 0.2 \times 0.6 + 0.75}\) | B1 |
| M1 | Attempt at \(P(B)\) or \(P(NB)\) anywhere involving sum of 2 or 3 products | |
| \(= \frac{0.12}{0.885}\) | A1 | Correct unsimplified num or denom of a fraction |
| \(= 0.136\ (8/59)\) | A1 [4] | Correct answer |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = 0.2$; $\mu = 96 \times 0.2 = 19.2$; $\sigma^2 = 96 \times 0.2 \times 0.8 = 15.36$ | B1 | $96 \times 0.2$ and $96 \times 0.2 \times 0.8$ seen |
| $P(<20) = P\!\left(z < \frac{19.5 - 19.2}{\sqrt{15.36}}\right) = P(z < 0.07654)$ | M1 | Standardising must have sq rt |
| | M1 | Continuity correction either 19.5 or 20.5 |
| | M1 | Correct area $(> 0.5)$ |
| $= 0.531$ | A1 [5] | Correct value |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(OT|B) = \frac{0.2 \times 0.6}{0.05 \times 0.3 + 0.2 \times 0.6 + 0.75}$ | B1 | Their $0.2 \times (0.6$ or $0.4)$ as numerator of a fraction |
| | M1 | Attempt at $P(B)$ or $P(NB)$ anywhere involving sum of 2 or 3 products |
| $= \frac{0.12}{0.885}$ | A1 | Correct unsimplified num or denom of a fraction |
| $= 0.136\ (8/59)$ | A1 [4] | Correct answer |
---6 Ana meets her friends once every day. For each day the probability that she is early is 0.05 and the probability that she is late is 0.75 . Otherwise she is on time.\\
(i) Find the probability that she is on time on fewer than 20 of the next 96 days.\\
(ii) If she is early there is a probability of 0.7 that she will eat a banana. If she is late she does not eat a banana. If she is on time there is a probability of 0.4 that she will eat a banana. Given that for one particular meeting with friends she does not eat a banana, find the probability that she is on time.
\hfill \mbox{\textit{CAIE S1 2012 Q6 [9]}}