| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Standard +0.3 This question tests standard application of linear combinations of normal variables with straightforward calculations. Part (i) requires finding the distribution of a sum of two identical normals, while part (ii) involves combining multiple normals with different parameters. Both are routine S2 procedures requiring no novel insight, though slightly above average difficulty due to the multi-variable setup in part (ii). |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(2R \sim N(13.0, 2 \times 0.23^2)\) | B1 |
| \(P(2R > 12.5) = 1 - \Phi\left(\frac{12.5-13}{\sqrt{0.1058}}\right)\) | M1 | Standardising and area > 0.5 |
| \(= \Phi(1.537) = 0.938\) | A1 [3] | Correct answer |
| (ii) | \(3R - B \sim N(8.2, 3 \times 0.23^2 + 0.46^2)\) | B1 |
| \(P((3R-B) > 6.7) = 1 - \Phi\left(\frac{6.7-8.2}{\sqrt{0.3703}}\right)\) | M1 | Considering \(P((3R-B) > 6.7)\) o.e. |
| \(= \Phi(2.4465) = 0.993\) | M1 A1 [4] | Correct probability area > 0.5. Correct answer |
(i) | $2R \sim N(13.0, 2 \times 0.23^2)$ | B1 | Correct mean and variance
$P(2R > 12.5) = 1 - \Phi\left(\frac{12.5-13}{\sqrt{0.1058}}\right)$ | M1 | Standardising and area > 0.5
$= \Phi(1.537) = 0.938$ | A1 [3] | Correct answer
(ii) | $3R - B \sim N(8.2, 3 \times 0.23^2 + 0.46^2)$ | B1 | Correct mean and variance
$P((3R-B) > 6.7) = 1 - \Phi\left(\frac{6.7-8.2}{\sqrt{0.3703}}\right)$ | M1 | Considering $P((3R-B) > 6.7)$ o.e.
$= \Phi(2.4465) = 0.993$ | M1 A1 [4] | Correct probability area > 0.5. Correct answer
---3 The lengths of red pencils are normally distributed with mean 6.5 cm and standard deviation 0.23 cm .\\
(i) Two red pencils are chosen at random. Find the probability that their total length is greater than 12.5 cm .
The lengths of black pencils are normally distributed with mean 11.3 cm and standard deviation 0.46 cm .\\
(ii) Find the probability that the total length of 3 red pencils is more than 6.7 cm greater than the length of 1 black pencil.
\hfill \mbox{\textit{CAIE S2 2008 Q3 [7]}}