| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pure expectation and variance calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard expectation and variance rules for linear combinations of independent random variables. The question requires only direct substitution into formulas E(aX+bY) = aE(X) + bE(Y) and Var(aX+bY) = a²Var(X) + b²Var(Y), with the additional recall that for Poisson distributions, variance equals mean. No problem-solving or conceptual insight is needed beyond memorized rules. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(E(2X-3Y) = 2E(X) - 3E(Y) = 16 - 18 = -2\) | M1 A1 | For multiplying by 2 and 3 resp and subt; For correct answer |
| 2 | ||
| (ii) \(\text{Var}(2X-3Y) = 4\text{Var}(X) + 9\text{Var}(Y) = 19.2 + 54 = 73.2\) | B1 M1 M1 A1 | For use of var\((Y) = 6\); For squaring 3 and 2; For adding variances (and nothing else); For correct final answer |
| 4 |
**(i)** $E(2X-3Y) = 2E(X) - 3E(Y) = 16 - 18 = -2$ | M1 A1 | For multiplying by 2 and 3 resp and subt; For correct answer
| | **2** | |
**(ii)** $\text{Var}(2X-3Y) = 4\text{Var}(X) + 9\text{Var}(Y) = 19.2 + 54 = 73.2$ | B1 M1 M1 A1 | For use of var$(Y) = 6$; For squaring 3 and 2; For adding variances (and nothing else); For correct final answer
| | **4** | |3 The independent random variables $X$ and $Y$ are such that $X$ has mean 8 and variance 4.8 and $Y$ has a Poisson distribution with mean 6. Find\\
(i) $\mathrm { E } ( 2 X - 3 Y )$,\\
(ii) $\operatorname { Var } ( 2 X - 3 Y )$.
\hfill \mbox{\textit{CAIE S2 2004 Q3 [6]}}