| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Same variable, two observations |
| Difficulty | Standard +0.3 This question tests standard applications of linear combinations of normal variables: part (i) requires finding the distribution of a sum (4 cans + box) and calculating a probability, while part (ii) involves the distribution of a difference between two independent normal variables. Both are routine applications of formulas typically taught in S2, requiring only substitution into variance rules and normal probability calculations without novel insight. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(W \sim N(2240, 848)\) | B2 | B1 each parameter |
| \(\frac{2200 - 2240}{\sqrt{848}} (= -1.374)\) | ||
| \(\Phi("-1.374") = 1 - \Phi("1.374") (= 0.0847)\) | M1A1 | Standardise either value and evaluate correctly |
| \(\frac{2300 - 2240}{\sqrt{848}} (= 2.060)\) | M1 | Correct combination of \(\Phi\)'s |
| \(\Phi("2.060") (= 0.9803)\) | ||
| \(\Phi("2.060") - [1 - \Phi("1.374")]\) | A1 | |
| \(= 0.896\) (3 sfs) | ||
| [6] | ||
| (ii) \(X_1 - X_2 \sim N(0, 392)\) | B1 | May be implied |
| \(\frac{20 - 0}{\sqrt{392}} (= 1.010)\) | M1 | |
| \((\Phi("1.010" = 0.8438)\) | ||
| \(P(X > 20) = 1 - \Phi("1.010") (= 0.1562)\) | A1 | |
| \(2 \times P(X > 20)\) | M1 | |
| \(= 0.312\) (3 sfs) | A1 | |
| [5] |
**(i)** $W \sim N(2240, 848)$ | B2 | B1 each parameter
$\frac{2200 - 2240}{\sqrt{848}} (= -1.374)$ | | |
$\Phi("-1.374") = 1 - \Phi("1.374") (= 0.0847)$ | M1A1 | Standardise either value and evaluate correctly
$\frac{2300 - 2240}{\sqrt{848}} (= 2.060)$ | M1 | Correct combination of $\Phi$'s
$\Phi("2.060") (= 0.9803)$ | | |
$\Phi("2.060") - [1 - \Phi("1.374")]$ | A1 |
$= 0.896$ (3 sfs) | |
| [6] |
**(ii)** $X_1 - X_2 \sim N(0, 392)$ | B1 | May be implied
$\frac{20 - 0}{\sqrt{392}} (= 1.010)$ | M1 |
$(\Phi("1.010" = 0.8438)$ | | |
$P(X > 20) = 1 - \Phi("1.010") (= 0.1562)$ | A1 |
$2 \times P(X > 20)$ | M1 |
$= 0.312$ (3 sfs) | A1 |
| [5] |5 Cans of drink are packed in boxes, each containing 4 cans. The weights of these cans are normally distributed with mean 510 g and standard deviation 14 g . The weights of the boxes, when empty, are independently normally distributed with mean 200 g and standard deviation 8 g .\\
(i) Find the probability that the total weight of a full box of cans is between 2200 g and 2300 g .\\
(ii) Two cans of drink are chosen at random. Find the probability that they differ in weight by more than 20 g .
\hfill \mbox{\textit{CAIE S2 2011 Q5 [11]}}