| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Normal approximation to binomial |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution and normal approximation. Part (i) is basic probability multiplication, part (ii) is standard binomial calculation with small n, and part (iii) is a routine normal approximation with continuity correction. All techniques are textbook exercises requiring no novel insight, though slightly easier than average due to clear setup and standard methods. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((0.05)(0.75)(0.15) = 0.00563\ (9/1600)\) | M1 | Multiplying 3 probs only, no Cs |
| B1 | 0.05 or 0.15 or \(\frac{1}{5} \times \frac{1}{4}\) seen | |
| A1 3 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{at least } 8) = P(8, 9, 10)\) | B1 | Binomial expression involving \((0.75)^r(0.25)^{10-r}\) and a C, \(r \neq 0\) or 10 |
| \(= {}_{10}C_8(0.75)^8(0.25)^2 + {}_{10}C_9(0.75)^9(0.25) + (0.75)^{10}\) | M1 | Correct unsimplified expression can be implied |
| \(= 0.526\) | A1 3 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mu = 90 \times 0.75 = 67.5\) | B1 | \(90 \times 0.75\ (67.5)\) and \(90 \times 0.75 \times 0.25\ (16.875\ \text{or}\ 16.9)\) seen |
| \(\sigma^2 = 90 \times 0.75 \times 0.25 = 16.875\) | ||
| \(P(X > 60) = 1 - \Phi\!\left(\dfrac{60.5 - 67.5}{\sqrt{16.875}}\right) = \Phi(1.704)\) | M1 | For standardising, with or without cc, must have \(\sqrt{\phantom{0}}\) on denom |
| M1 | For use of continuity correction 60.5 or 59.5 | |
| M1 | For finding an area \(> 0.5\) from their \(z\) | |
| \(= 0.956\) | A1 5 | For answer rounding to 0.956 |
## Question 7:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(0.05)(0.75)(0.15) = 0.00563\ (9/1600)$ | M1 | Multiplying 3 probs only, no Cs |
| | B1 | 0.05 or 0.15 or $\frac{1}{5} \times \frac{1}{4}$ seen |
| | A1 **3** | Correct answer |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{at least } 8) = P(8, 9, 10)$ | B1 | Binomial expression involving $(0.75)^r(0.25)^{10-r}$ and a C, $r \neq 0$ or 10 |
| $= {}_{10}C_8(0.75)^8(0.25)^2 + {}_{10}C_9(0.75)^9(0.25) + (0.75)^{10}$ | M1 | Correct unsimplified expression can be implied |
| $= 0.526$ | A1 **3** | Correct answer |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mu = 90 \times 0.75 = 67.5$ | B1 | $90 \times 0.75\ (67.5)$ and $90 \times 0.75 \times 0.25\ (16.875\ \text{or}\ 16.9)$ seen |
| $\sigma^2 = 90 \times 0.75 \times 0.25 = 16.875$ | | |
| $P(X > 60) = 1 - \Phi\!\left(\dfrac{60.5 - 67.5}{\sqrt{16.875}}\right) = \Phi(1.704)$ | M1 | For standardising, with or without cc, must have $\sqrt{\phantom{0}}$ on denom |
| | M1 | For use of continuity correction 60.5 or 59.5 |
| | M1 | For finding an area $> 0.5$ from their $z$ |
| $= 0.956$ | A1 **5** | For answer rounding to 0.956 |7 A die is biased so that the probability of throwing a 5 is 0.75 and the probabilities of throwing a 1,2 , 3 , 4 or 6 are all equal.\\
(i) The die is thrown three times. Find the probability that the result is a 1 followed by a 5 followed by any even number.\\
(ii) Find the probability that, out of 10 throws of this die, at least 8 throws result in a 5 .\\
(iii) The die is thrown 90 times. Using an appropriate approximation, find the probability that a 5 is thrown more than 60 times.
\hfill \mbox{\textit{CAIE S1 2008 Q7 [11]}}