| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | All components random including container |
| Difficulty | Challenging +1.2 Part (i) requires forming a linear combination of 7 independent normal variables (6 packets + 1 box), finding its mean and variance, then a standard probability calculation. Part (ii) is more challenging, requiring the insight to form P(X ≥ 8Y) as P(X - 8Y ≥ 0), then finding parameters of this new normal distribution. While the mechanics are A-level standard, the conceptual step of handling the inequality through a linear combination and the multi-stage reasoning across two non-trivial parts elevates this above average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6 \times 510 + 70 = (3130)\) | B1 | |
| \(6 \times 12^2 + 4^2 = (880)\) | B1 | |
| \(N(3130, 880)\) | ||
| \(\frac{3050-3130}{\sqrt{880}} = -2.697\) | M1 | Both. With their mean and variance (\(\geq 0\)). Allow without \(\sqrt{}\) |
| \(\frac{3150-3130}{\sqrt{880}} = 0.674\) | ||
| \(\Phi(0.674') - (1 - \Phi(2.697'))\) | M1 | Use of tables and attempt to find area consistent with their working |
| \(= 0.7499 - 0.0035\) | ||
| \(= 0.746\) (3 s.f.) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(510 - 8 \times 70 = (-50)\) | B1 | o.e. \(+50\); \(510/8\ -70\); \(-(510/8 - 70)\) |
| \(12^2 + 8^2 \times 4^2 = (1168)\) | B1 | o.e. \((12/8)^2 + 4^2\) |
| \(P - 8C \sim N(-50, 1168)\) | ||
| \(\frac{0-(-50)}{\sqrt{1168}} = 1.463\) | M1 | For standardising with attempt "\(P-8C\)" o.e with their mean and variance (\(\geq 0\)). Allow without \(\sqrt{}\) |
| \(1 - \Phi(1.463')\) | M1 | Use of tables and attempt to find area consistent with their working |
| \(= 0.0717\) (3 s.f.) | A1 [5] |
## Question 5:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6 \times 510 + 70 = (3130)$ | B1 | |
| $6 \times 12^2 + 4^2 = (880)$ | B1 | |
| $N(3130, 880)$ | | |
| $\frac{3050-3130}{\sqrt{880}} = -2.697$ | M1 | Both. With their mean and variance ($\geq 0$). Allow without $\sqrt{}$ |
| $\frac{3150-3130}{\sqrt{880}} = 0.674$ | | |
| $\Phi(0.674') - (1 - \Phi(2.697'))$ | M1 | Use of tables and attempt to find area consistent with their working |
| $= 0.7499 - 0.0035$ | | |
| $= 0.746$ (3 s.f.) | A1 [5] | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $510 - 8 \times 70 = (-50)$ | B1 | o.e. $+50$; $510/8\ -70$; $-(510/8 - 70)$ |
| $12^2 + 8^2 \times 4^2 = (1168)$ | B1 | o.e. $(12/8)^2 + 4^2$ |
| $P - 8C \sim N(-50, 1168)$ | | |
| $\frac{0-(-50)}{\sqrt{1168}} = 1.463$ | M1 | For standardising with attempt "$P-8C$" o.e with their mean and variance ($\geq 0$). Allow without $\sqrt{}$ |
| $1 - \Phi(1.463')$ | M1 | Use of tables and attempt to find area consistent with their working |
| $= 0.0717$ (3 s.f.) | A1 [5] | |5 Packets of cereal are packed in boxes, each containing 6 packets. The masses of the packets are normally distributed with mean 510 g and standard deviation 12 g . The masses of the empty boxes are normally distributed with mean 70 g and standard deviation 4 g .\\
(i) Find the probability that the total mass of a full box containing 6 packets is between 3050 g and 3150 g .\\
(ii) A packet and an empty box are chosen at random. Find the probability that the mass of the packet is at least 8 times the mass of the empty box.
\hfill \mbox{\textit{CAIE S2 2013 Q5 [10]}}