| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Challenging +1.2 This question requires understanding of linear combinations of normal random variables and sampling distributions, which are standard S2 topics. Part (i) involves forming a new random variable (difference of sums) and standardizing, while part (ii) applies the sampling distribution of means. Both parts are multi-step but follow well-established procedures without requiring novel insight—slightly above average difficulty due to the setup complexity and careful variance handling. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(T_1 + T_2 + T_4 - T_3 \sim N(-0.95, 4.345)\) | M1, B1, A1, M1 | Attempt to find mean and var of \(T_1 + T_2 + T_4 - T_3\) or; Correct mean (3.75 + 3.1 + 3.2 - 11); Correct variance; Finding P their [(\(T_1 + T_2 + T_4 - T_3\)) > 0] or |
| \(P\left[(T_1 + T_2 + T_4 - T_3) > 0\right] = P\left(z > \frac{0 - (-0.95)}{\sqrt{4.345}}\right) = P(z > 0.4557)\) | M1 | Standardising (appropriate variance involving all 4) and area < 0.5 |
| \(= 1 - 0.0.6758 = 0.324\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(\bar{X} \sim N(3.1, 0.785/6)\) | M1 | Normal distribution mean 3.1, var 0.785/6, can be implied OR N(18.6, 4.71) if working with totals |
| \(P(\bar{X} < 4) = P\left(z < \frac{4 - 3.1}{\sqrt{0.785/6}}\right) = P(z < 2.488)\) | M1 | Standardising with sq rt OR (24 – 18.6)/√4.71 no mixed methods |
| \(= 0.994\) | A1 | Correct answer |
(i) $T_1 + T_2 + T_4 - T_3 \sim N(-0.95, 4.345)$ | M1, B1, A1, M1 | Attempt to find mean and var of $T_1 + T_2 + T_4 - T_3$ or; Correct mean (3.75 + 3.1 + 3.2 - 11); Correct variance; Finding P their [($T_1 + T_2 + T_4 - T_3$) > 0] or
$P\left[(T_1 + T_2 + T_4 - T_3) > 0\right] = P\left(z > \frac{0 - (-0.95)}{\sqrt{4.345}}\right) = P(z > 0.4557)$ | M1 | Standardising (appropriate variance involving all 4) and area < 0.5
$= 1 - 0.0.6758 = 0.324$ | A1 | Correct answer
**[6]**
(ii) $\bar{X} \sim N(3.1, 0.785/6)$ | M1 | Normal distribution mean 3.1, var 0.785/6, can be implied OR N(18.6, 4.71) if working with totals
$P(\bar{X} < 4) = P\left(z < \frac{4 - 3.1}{\sqrt{0.785/6}}\right) = P(z < 2.488)$ | M1 | Standardising with sq rt OR (24 – 18.6)/√4.71 no mixed methods
$= 0.994$ | A1 | Correct answer
**[3]**6 When Sunil travels from his home in England to visit his relatives in India, his journey is in four stages. The times, in hours, for the stages have independent normal distributions as follows.
Bus from home to the airport: $\quad \mathrm { N } ( 3.75,1.45 )$\\
Waiting in the airport: $\quad \mathrm { N } ( 3.1,0.785 )$\\
Flight from England to India: $\quad \mathrm { N } ( 11,1.3 )$\\
Car in India to relatives: $\quad \mathrm { N } ( 3.2,0.81 )$\\
(i) Find the probability that the flight time is shorter than the total time for the other three stages.\\
(ii) Find the probability that, for 6 journeys to India, the mean time waiting in the airport is less than 4 hours.
\hfill \mbox{\textit{CAIE S2 2009 Q6 [9]}}