| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | November |
| Marks | 11 |
| 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 multi-part question testing basic probability concepts and normal approximation to binomial. Part (i) requires simple algebraic setup (2x + x = 1 for 3 odd and 3 even outcomes), part (ii) is direct binomial probability calculation, and part (iii) is a standard textbook application of normal approximation with continuity correction. All steps are routine with no novel insight required, making it 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/Working | Marks | Guidance |
| Let \(P(2,4,6)\) all \(= p\), then \(P(1,3,5)\) all \(= 2p\) | M1 | Using \(P(\text{even}) = 2P(\text{odd})\) or vice versa oe |
| \(3p + 6p = 1\) | M1 | Summing \(P(\text{odd} + \text{even})\) or \(P(1,2,3,4,5,6) = 1\) |
| \(p = 1/9\) so \(\text{prob}(3) = 2/9\ (0.222)\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(5,5,6) = \dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{1}{9} \times {}^3C_2\) | M1 | Multiply three probs together |
| \(= \dfrac{4}{243}\ (0.0165)\) | M1 | Multiply by 3, ie summing 3 options |
| A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mu = 100 \times \dfrac{1}{3} = 33.3,\ \sigma = 100 \times \dfrac{1}{3} \times \dfrac{2}{3} = 22.2\) | B1 | Unsimplified \(100/3\) and \(200/9\) seen |
| \(P(x \leqslant 37) = P\!\left(z \leqslant \dfrac{37.5 - \frac{100}{3}}{\sqrt{\frac{200}{9}}}\right) = P(z \leqslant 0.8839)\) | M1 | Standardising, need sq rt |
| M1 | \(36.5\) or \(37.5\) seen | |
| M1 | Correct area using their mean | |
| \(= 0.812\) | A1 [5] | Correct answer |
## Question 7 (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $P(2,4,6)$ all $= p$, then $P(1,3,5)$ all $= 2p$ | M1 | Using $P(\text{even}) = 2P(\text{odd})$ or vice versa oe |
| $3p + 6p = 1$ | M1 | Summing $P(\text{odd} + \text{even})$ or $P(1,2,3,4,5,6) = 1$ |
| $p = 1/9$ so $\text{prob}(3) = 2/9\ (0.222)$ | A1 [3] | Correct answer |
## Question 7 (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(5,5,6) = \dfrac{2}{9} \times \dfrac{2}{9} \times \dfrac{1}{9} \times {}^3C_2$ | M1 | Multiply three probs together |
| $= \dfrac{4}{243}\ (0.0165)$ | M1 | Multiply by 3, ie summing 3 options |
| | A1 [3] | Correct answer |
## Question 7 (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mu = 100 \times \dfrac{1}{3} = 33.3,\ \sigma = 100 \times \dfrac{1}{3} \times \dfrac{2}{3} = 22.2$ | B1 | Unsimplified $100/3$ and $200/9$ seen |
| $P(x \leqslant 37) = P\!\left(z \leqslant \dfrac{37.5 - \frac{100}{3}}{\sqrt{\frac{200}{9}}}\right) = P(z \leqslant 0.8839)$ | M1 | Standardising, need sq rt |
| | M1 | $36.5$ or $37.5$ seen |
| | M1 | Correct area using their mean |
| $= 0.812$ | A1 [5] | Correct answer |7 The faces of a biased die are numbered $1,2,3,4,5$ and 6 . The probabilities of throwing odd numbers are all the same. The probabilities of throwing even numbers are all the same. The probability of throwing an odd number is twice the probability of throwing an even number.\\
(i) Find the probability of throwing a 3 .\\
(ii) The die is thrown three times. Find the probability of throwing two 5 s and one 4 .\\
(iii) The die is thrown 100 times. Use an approximation to find the probability that an even number is thrown at most 37 times.
\hfill \mbox{\textit{CAIE S1 2015 Q7 [11]}}