| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum or total of normal variables |
| Difficulty | Moderate -0.3 This is a straightforward application of linear combinations of normal variables with clear setup: (i) requires finding P(sum > 5.95) using standard normal distribution properties, and (ii) extends this to sampling distribution of means. Both parts are direct textbook exercises requiring only formula application (sum of normals, CLT) with no conceptual challenges or problem-solving insight needed. Slightly easier than average due to explicit independence statement and routine calculations. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(T \sim N(1.54 \times 4, 0.05^2 \times 4)\) [\(\sim N(6.16, 0.01)\)] | B1 | For mult mean and variance by 4 |
| \(P(T>5.95) = 1 - \Phi\left(\frac{5.95-6.16}{\sqrt{0.01}}\right)\) | M1 | For standardising must have \(\sqrt{\phantom{0}}\) |
| \(= \Phi(2.1)\) | M1 | For correct area i.e. \(> 0.5\) |
| \(= 0.982\) | A1 (4) | For correct answer |
| (ii) \(Av \sim N(6.16, 0.01/20)\) | B1ft | For dividing their variance by 20 |
| \(P(Av > 6.2) = 1 - \Phi\left(\frac{6.2-6.16}{\sqrt{(0.01/20)}}\right)\) | M1 | For standardising (must use consistent values) |
| \(= 1 - \Phi(1.789)\) | ||
| \(= 1 - 0.9633\) | ||
| \(= 0.0367\) or \(0.0368\) | A1 (3) | For correct answer |
| or Tot \(\sim N(123.2, 0.2)\) | B1ft | |
| \(P(\text{Tot}>124) = 1-\Phi\left(\frac{123.2-124}{\sqrt{0.2}}\right)\) etc. | M1 |
**(i)** $T \sim N(1.54 \times 4, 0.05^2 \times 4)$ [$\sim N(6.16, 0.01)$] | B1 | For mult mean and variance by 4
$P(T>5.95) = 1 - \Phi\left(\frac{5.95-6.16}{\sqrt{0.01}}\right)$ | M1 | For standardising must have $\sqrt{\phantom{0}}$
$= \Phi(2.1)$ | M1 | For correct area i.e. $> 0.5$
$= 0.982$ | A1 (4) | For correct answer
**(ii)** $Av \sim N(6.16, 0.01/20)$ | B1ft | For dividing their variance by 20
$P(Av > 6.2) = 1 - \Phi\left(\frac{6.2-6.16}{\sqrt{(0.01/20)}}\right)$ | M1 | For standardising (must use consistent values)
$= 1 - \Phi(1.789)$ | |
$= 1 - 0.9633$ | |
$= 0.0367$ or $0.0368$ | A1 (3) | For correct answer
or Tot $\sim N(123.2, 0.2)$ | B1ft |
$P(\text{Tot}>124) = 1-\Phi\left(\frac{123.2-124}{\sqrt{0.2}}\right)$ etc. | M1 |5 A clock contains 4 new batteries each of which gives a voltage which is normally distributed with mean 1.54 volts and standard deviation 0.05 volts. The voltages of the batteries are independent. The clock will only work if the total voltage is greater than 5.95 volts.\\
(i) Find the probability that the clock will work.\\
(ii) Find the probability that the average total voltage of the batteries of 20 clocks chosen at random exceeds 6.2 volts.
\hfill \mbox{\textit{CAIE S2 2005 Q5 [7]}}