| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2002 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal variables with standard formulas. Part (i) requires finding the distribution of 12X + Y (sum of bottles plus rack), then a normal probability calculation. Part (ii) needs the distribution of X₁ - X₂ (difference of two bottles). Both parts are textbook exercises requiring only formula recall and basic normal table usage, making it slightly easier than average for A-level statistics. |
| 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) \(W \sim N(17.6, 0.133(2))\) | B1, B1 | For correct mean. For correct variance. |
| \(\Phi\left(\frac{18 - 17.6}{\sqrt{0.01332}}\right) = (0.8633)\) | M1 | For standardising and using tables. |
| \(\Phi\left(\frac{17 - 17.6}{\sqrt{0.01332}}\right) = 1 - 0.9499 = (0.0501)\) | M1 | For standardising and using tables. |
| \(0.8633 - 0.0501 = 0.813\) | A1 (5 marks) | For correct answer. |
| (ii) Wt diff \(D \sim N(0, 0.0072)\) | B1 | For correct mean and variance. |
| \(P(D > 0.05) = 1 - \Phi\left(\frac{0.05}{\sqrt{0.0072}}\right) = 1 - \Phi(0.589) = 1 - 0.278 = 0.278\) | M1, A1 | For standardising and using tables. For \(0.278\) (could be implied). |
| \(P(D < 0.05) = 0.278\) | M1 | For finding the other probability. |
| \(0.278 + 0.278 = 0.556\) | A1 (5 marks) | For correct answer. |
(i) $W \sim N(17.6, 0.133(2))$ | B1, B1 | For correct mean. For correct variance.
$\Phi\left(\frac{18 - 17.6}{\sqrt{0.01332}}\right) = (0.8633)$ | M1 | For standardising and using tables.
$\Phi\left(\frac{17 - 17.6}{\sqrt{0.01332}}\right) = 1 - 0.9499 = (0.0501)$ | M1 | For standardising and using tables.
$0.8633 - 0.0501 = 0.813$ | A1 (5 marks) | For correct answer.
(ii) Wt diff $D \sim N(0, 0.0072)$ | B1 | For correct mean and variance.
$P(D > 0.05) = 1 - \Phi\left(\frac{0.05}{\sqrt{0.0072}}\right) = 1 - \Phi(0.589) = 1 - 0.278 = 0.278$ | M1, A1 | For standardising and using tables. For $0.278$ (could be implied).
$P(D < 0.05) = 0.278$ | M1 | For finding the other probability.
$0.278 + 0.278 = 0.556$ | A1 (5 marks) | For correct answer.Bottles of wine are stacked in racks of 12. The weights of these bottles are normally distributed with mean 1.3 kg and standard deviation 0.06 kg. The weights of the empty racks are normally distributed with mean 2 kg and standard deviation 0.3 kg.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the total weight of a full rack of 12 bottles of wine is between 17 kg and 18 kg. [5]
\item Two bottles of wine are chosen at random. Find the probability that they differ in weight by more than 0.05 kg. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2002 Q7 [10]}}