| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum versus sum comparison |
| Difficulty | Standard +0.8 This question requires understanding of linear combinations of independent normal random variables, including forming new distributions from sums and scaled variables. Part (i) involves comparing 3X vs 2Y (requiring variance calculations for sums), and part (ii) involves 2X + 0.5Y. While the underlying theory is A-level appropriate, correctly setting up the distributions, calculating combined variances, and standardizing requires careful multi-step work beyond routine application, 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) \(3C \sim N(990, 5.2^2 \times 3) = (N(990, 81.12))\) | B1 | Correct mean for both 3 cans cola and 2 bottles water |
| \(2W \sim N(1000, 7.1^2 \times 2 = (N(1000, 100.82))\) | B1 | Correct variance for both 3 cans cola and 2 bottles water |
| \(3C - 2W \sim N(-10, 181.94)\) | M1 | Correct method for mean and variance for \(3C - 2W\) or vice versa |
| \(P(3C - 2W < 0) = \Phi\left(\frac{0 - (-10)}{\sqrt{181.94}}\right)\) | M1 | Standardising and using tables, need the sq root and area \(> 0.5\) |
| \(= \Phi(0.741) = 0.771\) | A1 [5] | Correct answer |
| (ii) new drink \(\sim N(910, 2 \times 5.2^2 + 0.5^2 \times 7.1^2)\) | B1 | Correct mean for new drink |
| \(\sim N(910, 66.68)\) | B1 | Correct variance for new drink |
| \(P(ND > 900) = 1 - P\left(z < \frac{900 - 910}{\sqrt{66.68}}\right)\) | M1 | Standardising with sq rt and using tables |
| \(= 1 - P(z < -1.225) = \Phi(1.225) = 0.8897 (0.890)\) | A1 [4] | Correct answer |
**(i)** $3C \sim N(990, 5.2^2 \times 3) = (N(990, 81.12))$ | B1 | Correct mean for both 3 cans cola and 2 bottles water
$2W \sim N(1000, 7.1^2 \times 2 = (N(1000, 100.82))$ | B1 | Correct variance for both 3 cans cola and 2 bottles water
$3C - 2W \sim N(-10, 181.94)$ | M1 | Correct method for mean and variance for $3C - 2W$ or vice versa
$P(3C - 2W < 0) = \Phi\left(\frac{0 - (-10)}{\sqrt{181.94}}\right)$ | M1 | Standardising and using tables, need the sq root and area $> 0.5$
$= \Phi(0.741) = 0.771$ | A1 [5] | Correct answer
**(ii)** new drink $\sim N(910, 2 \times 5.2^2 + 0.5^2 \times 7.1^2)$ | B1 | Correct mean for new drink
$\sim N(910, 66.68)$ | B1 | Correct variance for new drink
$P(ND > 900) = 1 - P\left(z < \frac{900 - 910}{\sqrt{66.68}}\right)$ | M1 | Standardising with sq rt and using tables
$= 1 - P(z < -1.225) = \Phi(1.225) = 0.8897 (0.890)$ | A1 [4] | Correct answer7 The volume of liquid in cans of cola is normally distributed with mean 330 millilitres and standard deviation 5.2 millilitres. The volume of liquid in bottles of tonic water is normally distributed with mean 500 millilitres and standard deviation 7.1 millilitres.\\
(i) Find the probability that 3 randomly chosen cans of cola contain less liquid than 2 randomly chosen bottles of tonic water.\\
(ii) A new drink is made by mixing the contents of 2 cans of cola with half a bottle of tonic water. Find the probability that the volume of the new drink is more than 900 millilitres.
\hfill \mbox{\textit{CAIE S2 2009 Q7 [9]}}