| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| 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 Part (i) is a straightforward binomial probability calculation with small n requiring P(X<3). Part (ii) involves recognizing when to apply normal approximation to binomial (n=125 is large), calculating parameters correctly (p=0.64 for cars), and applying continuity correction. Both parts are standard textbook exercises testing routine application of techniques with no novel problem-solving required, making this slightly easier than average. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X < 3) = P(0) + P(1) + P(2)\) | M1 | Binomial term with \(^{11}C_r p^r (1-p)^{11-r}\) seen |
| \(= (0.84)^{11} + (0.16)(0.84)^{10} \times ^{11}C_1 + (0.16)^2(0.84)^9 \times ^{11}C_2\) | M1 | Correct expression for \(P(0,1,2)\) or \(P(0,1,2,3)\); can have wrong \(p\) |
| \(= 0.1469 + 0.30782 + 0.2931\) | ||
| \(= 0.748\) | A1 [3] | Correct final answer. Normal approx M0 M0 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mu = 125 \times 0.64 = 80\) | B1 | 80 and 28.8 or 5.37 seen |
| \(\sigma^2 = 125 \times 0.64 \times 0.36 = 28.8\) | ||
| \(P(X > 73) = 1 - \Phi\left(\frac{73.5 - 80}{\sqrt{28.8}}\right)\) | M1 | Standardising, with or without cc, must have sq rt in denom |
| M1 | Continuity correction 73.5 or 72.5 only | |
| \(= \Phi(1.211)\) | M1 | Correct region (\(> 0.5\) if mean \(> 73.5\), vv if mean \(< 73.5\)) |
| \(= 0.887\) | A1 [5] | Correct answer |
## Question 3:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X < 3) = P(0) + P(1) + P(2)$ | M1 | Binomial term with $^{11}C_r p^r (1-p)^{11-r}$ seen |
| $= (0.84)^{11} + (0.16)(0.84)^{10} \times ^{11}C_1 + (0.16)^2(0.84)^9 \times ^{11}C_2$ | M1 | Correct expression for $P(0,1,2)$ or $P(0,1,2,3)$; can have wrong $p$ |
| $= 0.1469 + 0.30782 + 0.2931$ | | |
| $= 0.748$ | A1 **[3]** | Correct final answer. Normal approx M0 M0 A0 |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mu = 125 \times 0.64 = 80$ | B1 | 80 and 28.8 or 5.37 seen |
| $\sigma^2 = 125 \times 0.64 \times 0.36 = 28.8$ | | |
| $P(X > 73) = 1 - \Phi\left(\frac{73.5 - 80}{\sqrt{28.8}}\right)$ | M1 | Standardising, with or without cc, must have sq rt in denom |
| | M1 | Continuity correction 73.5 or 72.5 only |
| $= \Phi(1.211)$ | M1 | Correct region ($> 0.5$ if mean $> 73.5$, vv if mean $< 73.5$) |
| $= 0.887$ | A1 **[5]** | Correct answer |
---3 On a certain road $20 \%$ of the vehicles are trucks, $16 \%$ are buses and the remainder are cars.\\
(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses.\\
(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the probability that more than 73 are cars.
\hfill \mbox{\textit{CAIE S1 2009 Q3 [8]}}