| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Total journey time probabilities |
| Difficulty | Standard +0.3 This question tests standard linear combinations of normal distributions with straightforward applications: (i) requires finding the distribution of a sum of independent normals over 5 days and a single probability calculation, (ii) requires finding the distribution of a difference. Both are direct applications of well-rehearsed techniques with no conceptual surprises, making it slightly easier than average for an A-level statistics question. |
| 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) \(5 \times (22.4 + 20.3)\) & \(5 \times (4.8^2 + 5.2^2)\) | B1 B1 | For correct expression for new mean; For correct expression for new variance |
| \((= 213.5)\) | \((= 250.4)\) | |
| \(z = \frac{180 - \text{"213.5"}}{\sqrt{\text{"250.4"}}} (= -2.117)\) | M1 | Standardising and use of tables |
| \(1 - \Phi(\text{"-2.117"}) = \Phi(\text{"2.117"}) = 0.983\) (3 sfs) | A1 [4] | (no sd/var mixes + no cc) |
| (ii) \(P(H - W > 0)\) or \(P(W - H < 0)\) | M1 | Or \(\pm 2.1\) |
| \(20.3 - 22.4\) & \(4.8^2 + 5.2^2\) | B1 B1 | Correct expression for new mean; Correct expression for new variance |
| \((= -2.1)\) | \((= 50.08)\) | |
| \(z = \frac{0 - (-2.1)}{\sqrt{50.08}} (= 0.297)\) | M1 | Standardising and using tables |
| \(1 - \Phi(\text{"-0.297"}) (= 1 - 0.6168) = 0.383\) (3 sfs) | A1 [5] | (no sd/var mixes + no cc) |
**(i)** $5 \times (22.4 + 20.3)$ & $5 \times (4.8^2 + 5.2^2)$ | B1 B1 | For correct expression for new mean; For correct expression for new variance
$(= 213.5)$ | | $(= 250.4)$
$z = \frac{180 - \text{"213.5"}}{\sqrt{\text{"250.4"}}} (= -2.117)$ | M1 | Standardising and use of tables
$1 - \Phi(\text{"-2.117"}) = \Phi(\text{"2.117"}) = 0.983$ (3 sfs) | A1 [4] | (no sd/var mixes + no cc)
**(ii)** $P(H - W > 0)$ or $P(W - H < 0)$ | M1 | Or $\pm 2.1$
$20.3 - 22.4$ & $4.8^2 + 5.2^2$ | B1 B1 | Correct expression for new mean; Correct expression for new variance
$(= -2.1)$ | | $(= 50.08)$
$z = \frac{0 - (-2.1)}{\sqrt{50.08}} (= 0.297)$ | M1 | Standardising and using tables
$1 - \Phi(\text{"-0.297"}) (= 1 - 0.6168) = 0.383$ (3 sfs) | A1 [5] | (no sd/var mixes + no cc)6 Yu Ming travels to work and returns home once each day. The times, in minutes, that he takes to travel to work and to return home are represented by the independent random variables $W$ and $H$ with distributions $\mathrm { N } \left( 22.4,4.8 ^ { 2 } \right)$ and $\mathrm { N } \left( 20.3,5.2 ^ { 2 } \right)$ respectively.\\
(i) Find the probability that Yu Ming's total travelling time during a 5-day period is greater than 180 minutes.\\
(ii) Find the probability that, on a particular day, Yu Ming takes longer to return home than he takes to travel to work.
\hfill \mbox{\textit{CAIE S2 2010 Q6 [9]}}