| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Challenging +1.2 This is a multi-part question on linear combinations of normal random variables requiring understanding of variance properties and independence. Part (i) is standard (sum of independent normals), part (ii) tests understanding that doubling quantity doubles mean and quadruples variance, and part (iii) requires finding the distribution of a difference of two independent normal variables. While it has multiple stages and requires careful bookkeeping of components, the techniques are all standard S2 content with no novel insight required—just systematic application of learned rules. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sim N(157, 109.47)\) | M1, B1, B1 | For summing means of 2 trips + fuel + pay and variances of 2 trips + fuel; Correct mean; Correct variance |
| \(P(T_1 < 154) = \Phi\left(\frac{154-157}{\sqrt{109.47}}\right) = \Phi(-0.2867) = 1 - 0.6130 = 0.387\) | M1, A1 (5 marks total) | For standardising, can have cc, no sq rt; For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Variance = \(1.7^2 \times 4 = 11.56\) | B1, B1 (2 marks) | Correct mean; Correct variance |
| (iii) Total car 2, \(T_2 \sim N(69 \times 2 + 10 + 4, 5.2^2 \times 2 + 11.56) \sim N(152, 65.64)\) | B1ft, B1ft | Correct mean, ft on their (ii); Correct variance, ft on their (ii) |
| \(T_1 - T_2 \sim N(5, 175.11)\) | M1 | For considering \(P(T_1 - T_2 > 0)\) or equivalent |
| \(P(T_1 - T_2 > 0) = 1 - \Phi\left((0-5)\sqrt{175.11}\right) = \Phi(0.378) = 0.647\) | M1, A1 (5 marks total) | For standardising and using tables; For correct answer |
**(i)** Total time $T_1 \sim N(74 \times 2 + 5 + 4, 7.3^2 \times 2 + 1.7^2)$
$\sim N(157, 109.47)$ | M1, B1, B1 | For summing means of 2 trips + fuel + pay and variances of 2 trips + fuel; Correct mean; Correct variance
$P(T_1 < 154) = \Phi\left(\frac{154-157}{\sqrt{109.47}}\right) = \Phi(-0.2867) = 1 - 0.6130 = 0.387$ | M1, A1 (5 marks total) | For standardising, can have cc, no sq rt; For correct answer
**(ii)** Mean = 10
Variance = $1.7^2 \times 4 = 11.56$ | B1, B1 (2 marks) | Correct mean; Correct variance
**(iii)** Total car 2, $T_2 \sim N(69 \times 2 + 10 + 4, 5.2^2 \times 2 + 11.56) \sim N(152, 65.64)$ | B1ft, B1ft | Correct mean, ft on their (ii); Correct variance, ft on their (ii)
$T_1 - T_2 \sim N(5, 175.11)$ | M1 | For considering $P(T_1 - T_2 > 0)$ or equivalent
$P(T_1 - T_2 > 0) = 1 - \Phi\left((0-5)\sqrt{175.11}\right) = \Phi(0.378) = 0.647$ | M1, A1 (5 marks total) | For standardising and using tables; For correct answer7 A journey in a certain car consists of two stages with a stop for filling up with fuel after the first stage. The length of time, $T$ minutes, taken for each stage has a normal distribution with mean 74 and standard deviation 7.3. The length of time, $F$ minutes, it takes to fill up with fuel has a normal distribution with mean 5 and standard deviation 1.7. The length of time it takes to pay for the fuel is exactly 4 minutes. The variables $T$ and $F$ are independent and the times for the two stages are independent of each other.\\
(i) Find the probability that the total time for the journey is less than 154 minutes.\\
(ii) A second car has a fuel tank with exactly twice the capacity of the first car. Find the mean and variance of this car's fuel fill-up time.\\
(iii) This second car's time for each stage of the journey follows a normal distribution with mean 69 minutes and standard deviation 5.2 minutes. The length of time it takes to pay for the fuel for this car is also exactly 4 minutes. Find the probability that the total time for the journey taken by the first car is more than the total time taken by the second car.
\hfill \mbox{\textit{CAIE S2 2005 Q7 [12]}}