| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | At least one success |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question with three standard parts: (i) direct binomial probability calculation, (ii) finding sample size using 'at least one' complement (1 - 0.8^n ≥ 0.85), and (iii) normal approximation for large n. All parts use routine techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(0) = (0.8)^{15}\ (= 0.03518)\) | B1 | For correct numerical expression for \(P(0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(2) = {}_{15}C_2\times(0.2)^2\times(0.8)^{13}\ (= 0.2309)\) | B1 | For correct numerical expression for \(P(1)\) or \(P(2)\) |
| \(P(X\leq2) = 0.398\) | B1 | For answer rounding to 0.398 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1-(0.8)^n \geq 0.85\) | M1 | For an equality/inequality involving \(0.8,\ n,\ 0.85\) |
| \(0.15\geq(0.8)^n\) | M1dep | For solving attempt (could be trial and error or lg) |
| \(n=9\) | A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sigma^2 = 1600\times0.2\times0.8 = 256\) | B1 | For both mean and variance correct |
| \(P(X\geq290)\) or \(P(X<350)\) | M1 | For standardising, with or without cc; must have \(\sqrt{\ }\) on denom |
| \(= 1-\Phi\!\left(\frac{289.5-320}{\sqrt{256}}\right) = 1-\Phi(-1.906)\) | M1 | For use of continuity correction 289.5 or 290.5 |
| \(= \Phi(1.906) = 0.972\) | M1 | For finding an area \(> 0.5\) from their \(z\) |
| A1 | For answer rounding to 0.972 |
# Question 7:
## Part (i)
$P(0) = (0.8)^{15}\ (= 0.03518)$ | B1 | For correct numerical expression for $P(0)$
$P(1) = {}_{15}C_1\times(0.2)\times(0.8)^{14}\ (= 0.1319)$
$P(2) = {}_{15}C_2\times(0.2)^2\times(0.8)^{13}\ (= 0.2309)$ | B1 | For correct numerical expression for $P(1)$ or $P(2)$
$P(X\leq2) = 0.398$ | B1 | For answer rounding to 0.398
**Total: 3**
## Part (ii)
$1-(0.8)^n \geq 0.85$ | M1 | For an equality/inequality involving $0.8,\ n,\ 0.85$
$0.15\geq(0.8)^n$ | M1dep | For solving attempt (could be trial and error or lg)
$n=9$ | A1 | For correct answer
**Total: 3**
## Part (iii)
$\mu = 1600\times0.2 = 320$
$\sigma^2 = 1600\times0.2\times0.8 = 256$ | B1 | For both mean and variance correct
$P(X\geq290)$ or $P(X<350)$ | M1 | For standardising, with or without cc; must have $\sqrt{\ }$ on denom
$= 1-\Phi\!\left(\frac{289.5-320}{\sqrt{256}}\right) = 1-\Phi(-1.906)$ | M1 | For use of continuity correction 289.5 or 290.5
$= \Phi(1.906) = 0.972$ | M1 | For finding an area $> 0.5$ from their $z$
| A1 | For answer rounding to 0.972
**Total: 5**7 A shop sells old video tapes, of which 1 in 5 on average are known to be damaged.\\
(i) A random sample of 15 tapes is taken. Find the probability that at most 2 are damaged.\\
(ii) Find the smallest value of $n$ if there is a probability of at least 0.85 that a random sample of $n$ tapes contains at least one damaged tape.\\
(iii) A random sample of 1600 tapes is taken. Use a suitable approximation to find the probability that there are at least 290 damaged tapes.
\hfill \mbox{\textit{CAIE S1 2004 Q7 [11]}}