| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Overbooking probability problems |
| Difficulty | Standard +0.3 This is a standard application of normal approximation to binomial distribution with straightforward setup. Part (i) requires recognizing that 'more than 210 arrive' means 'fewer than 3 don't arrive', then applying continuity correction. Part (ii) combines two independent binomial distributions. While multi-step, both parts follow textbook procedures without requiring novel insight, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Poisson mean 4.26 \(P(\text{overbooked}) = P(0, 1, 2 \text{ not arrive}) = e^{-4.26}\left(1 + 4.26 + \frac{4.26^2}{2}\right) = 0.202\) | B1, M1, A1 ft, A1 [4] | Correct mean. Poisson attempt at P(0,1,2), their 4.26, allow end errors. Correct unsimplified answer, ft their mean. Correct answer, as final answer. |
| (ii) mean \(= 135/75 = 1.8\) \(\sum(\text{Poissons}) \text{ new mean} = 4.26 + 1.8 = 6.06\) \(P(5) = e^{-6.06}\left(\frac{6.06^2}{5!}\right) = 0.159\) | M1, M1, A1 [3] | Adding their two means. Attempt at Poisson P(5) with their mean. Correct answer |
| (i) SR Normal: \(B1\) \(N(4.26, 4.17(5))\) or \(N(208.74 (209), 4.17(5))\) | \(B1\) | Prob \(= 0.195\) |
| SR Binomial: \(B(213, 1/50)\) | \(B1\) | correct unsimplified expression for P(0,1,2) |
| \(B1\) | Prob \(= 0.200\) or \(0.199\) | |
| (ii) SR Binomials | M2 | all cases of binomial products |
| A1 | prob \(= 0.160\) |
(i) Poisson mean 4.26 $P(\text{overbooked}) = P(0, 1, 2 \text{ not arrive}) = e^{-4.26}\left(1 + 4.26 + \frac{4.26^2}{2}\right) = 0.202$ | B1, M1, A1 ft, A1 [4] | Correct mean. Poisson attempt at P(0,1,2), their 4.26, allow end errors. Correct unsimplified answer, ft their mean. Correct answer, as final answer.
(ii) mean $= 135/75 = 1.8$ $\sum(\text{Poissons}) \text{ new mean} = 4.26 + 1.8 = 6.06$ $P(5) = e^{-6.06}\left(\frac{6.06^2}{5!}\right) = 0.159$ | M1, M1, A1 [3] | Adding their two means. Attempt at Poisson P(5) with their mean. Correct answer
(i) SR Normal: $B1$ $N(4.26, 4.17(5))$ or $N(208.74 (209), 4.17(5))$ | $B1$ | Prob $= 0.195$
SR Binomial: $B(213, 1/50)$ | $B1$ | correct unsimplified expression for P(0,1,2)
$B1$ | Prob $= 0.200$ or $0.199$
(ii) SR Binomials | M2 | all cases of binomial products
| A1 | prob $= 0.160$3 An airline knows that some people who have bought tickets may not arrive for the flight. The airline therefore sells more tickets than the number of seats that are available. For one flight there are 210 seats available and 213 people have bought tickets. The probability of any person who has bought a ticket not arriving for the flight is $\frac { 1 } { 50 }$.\\
(i) By considering the number of people who do not arrive for the flight, use a suitable approximation to calculate the probability that more people will arrive than there are seats available.
Independently, on another flight for which 135 people have bought tickets, the probability of any person not arriving is $\frac { 1 } { 75 }$.\\
(ii) Calculate the probability that, for both these flights, the total number of people who do not arrive is 5 .
\hfill \mbox{\textit{CAIE S2 2009 Q3 [7]}}