| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for linear combinations of independent normal variables. Part (i) is routine verification using E(aX+bY)=aE(X)+bE(Y) and Var(aX+bY)=a²Var(X)+b²Var(Y). Part (ii) requires the same techniques plus a standard normal probability calculation. While it involves multiple steps, each step follows directly from learned formulas with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5M + 2W \sim N(355 + 114,\ 7^2 \times 5 + 5^2 \times 2) \sim N(469, 295)\) | B1 | For mean \(= 5\times71 + 2\times57\) |
| B1 2 | For variance \(= 7^2\times5 + 5^2\times2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim 4M + 3W \sim N(455, 271)\) | B1 | For correct mean and variance of \(4M+3W\) |
| \(X - Y \sim (5M+2W)-(4M+3W) \sim N(14, 566)\) | M1 | For adding their two variances and subtracting their two means |
| Mean \(= 14\), s.d. \(= \sqrt{566} = 23.8\) | A1ft | For both correct (must be s.d.), ft on wrong mean and var of \(Y\) |
| \(P(X - Y > 22) = 1 - \Phi\!\left(\frac{22-14}{\sqrt{566}}\right)\) | M1 | For standardising and using tables, either end; need the sq rt |
| \(= 1 - \Phi(0.3363)\) \(= 1 - 0.631\ \text{or}\ 1 - 0.632\) \(= 0.368\ \text{or}\ 0.369\) | A1 5 | For correct answer |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5M + 2W \sim N(355 + 114,\ 7^2 \times 5 + 5^2 \times 2) \sim N(469, 295)$ | B1 | For mean $= 5\times71 + 2\times57$ |
| | B1 **2** | For variance $= 7^2\times5 + 5^2\times2$ |
## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim 4M + 3W \sim N(455, 271)$ | B1 | For correct mean and variance of $4M+3W$ |
| $X - Y \sim (5M+2W)-(4M+3W) \sim N(14, 566)$ | M1 | For adding their two variances and subtracting their two means |
| Mean $= 14$, s.d. $= \sqrt{566} = 23.8$ | A1ft | For both correct (must be s.d.), ft on wrong mean and var of $Y$ |
| $P(X - Y > 22) = 1 - \Phi\!\left(\frac{22-14}{\sqrt{566}}\right)$ | M1 | For standardising and using tables, either end; need the sq rt |
| $= 1 - \Phi(0.3363)$ $= 1 - 0.631\ \text{or}\ 1 - 0.632$ $= 0.368\ \text{or}\ 0.369$ | A1 **5** | For correct answer |4 The weights of men follow a normal distribution with mean 71 kg and standard deviation 7 kg . The weights of women follow a normal distribution with mean 57 kg and standard deviation 5 kg . The total weight of 5 men and 2 women chosen randomly is denoted by $X \mathrm {~kg}$.\\
(i) Show that $\mathrm { E } ( X ) = 469$ and $\operatorname { Var } ( X ) = 295$.\\
(ii) The total weight of 4 men and 3 women chosen randomly is denoted by $Y \mathrm {~kg}$. Find the mean and standard deviation of $X - Y$ and hence find $\mathrm { P } ( X - Y > 22 )$.
\hfill \mbox{\textit{CAIE S2 2004 Q4 [7]}}