| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Standard +0.8 Part (i) requires forming the linear combination X - 5Y and finding P(X - 5Y > 0), which demands understanding of how to combine independent normal variables with a non-trivial coefficient. Part (ii) involves unit conversion and applying variance rules. This goes beyond routine application of formulas, requiring careful setup and interpretation, placing it moderately above average difficulty. |
| 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) \(Mr - 5Mrs \sim N(512 - 5 \times 89, 62^2 + 25 \times 7.4^2) \sim N(67, 5213)\) | B1, B1 | Correct unsimplified mean; Correct unsimplified variance |
| \(P(Mr > 5 Mrs) = P(Mr - 5 Mrs > 0) = P\left(z > \frac{0-67}{\sqrt{5213}}\right) = P(z > -0.9280) = 0.823\) | M1, M1, A1 [5] | Using distribution \(Mr - 5 Mrs\); Standardising and using tables; Correct answer |
| (ii) \(Mr + Mrs \sim N(601, 62^2 + 7.4^2)\) | B1 | Correct mean and variance |
| \(E[5/8(Mr + Mrs)] = 376\) miles | B1 | Correct answer |
| \(Var[5/8(Mr + Mrs)] = \frac{25}{64} \times 3898.76 = 1520\) | B1 [3] | SR Two separate answers 320 and 55.6 B1 |
**(i)** $Mr - 5Mrs \sim N(512 - 5 \times 89, 62^2 + 25 \times 7.4^2) \sim N(67, 5213)$ | B1, B1 | Correct unsimplified mean; Correct unsimplified variance
$P(Mr > 5 Mrs) = P(Mr - 5 Mrs > 0) = P\left(z > \frac{0-67}{\sqrt{5213}}\right) = P(z > -0.9280) = 0.823$ | M1, M1, A1 [5] | Using distribution $Mr - 5 Mrs$; Standardising and using tables; Correct answer
**(ii)** $Mr + Mrs \sim N(601, 62^2 + 7.4^2)$ | B1 | Correct mean and variance
$E[5/8(Mr + Mrs)] = 376$ miles | B1 | Correct answer
$Var[5/8(Mr + Mrs)] = \frac{25}{64} \times 3898.76 = 1520$ | B1 [3] | SR Two separate answers 320 and 55.6 B1
$sd = 39.0$ miles4 The weekly distance in kilometres driven by Mr Parry has a normal distribution with mean 512 and standard deviation 62. Independently, the weekly distance in kilometres driven by Mrs Parry has a normal distribution with mean 89 and standard deviation 7.4.\\
(i) Find the probability that, in a randomly chosen week, Mr Parry drives more than 5 times as far as Mrs Parry.\\
(ii) Find the mean and standard deviation of the total of the weekly distances in miles driven by Mr Parry and Mrs Parry. Use the approximation 8 kilometres $= 5$ miles.
\hfill \mbox{\textit{CAIE S2 2010 Q4 [8]}}