| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 9 |
| 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 question tests standard linear combinations of normal distributions with straightforward application of formulas. Part (i) requires finding the distribution of 9M + 7W and calculating P(total < 1200), while part (ii) involves finding P(W > M) = P(W - M > 0). Both are routine applications of the key result that linear combinations of independent normals are normal, requiring only careful arithmetic with means and variances—no novel insight or complex problem-solving needed. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(T) = 9 \times 78 + 7 \times 66 \quad (= 1164)\) | B1 | Or \(9 \times 78 + 7 \times 66 - 1200\) |
| \(\text{Var}(T) = 9 \times 7^2 + 7 \times 5^2 \quad (= 616)\) | B1 | |
| \(\dfrac{1200 - \text{'1164'}}{\sqrt{\text{'616'}}} \quad (= 1.450)\) | M1 | \(\pm\) Allow without \(\sqrt{}\) |
| \(P(z < 1.450) = \Phi(1.450)\) | M1 | |
| \(= 0.927\) (3 sf) | A1 | Correct tail consistent with their mean |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(D) = 66 - 78 \quad (= -12)\) | B1 | Both needed |
| \(\text{Var}(D) = 7^2 + 5^2 \quad (= 74)\) | ||
| \(\dfrac{0 - (\text{'-12'})}{\sqrt{74}} \quad (= 1.395)\) | M1 | \(\pm\) Allow without \(\sqrt{}\) |
| \(P(D > 0) = 1 - \Phi(\text{'1.395'})\) | M1 | Correct tail consistent with their mean |
| \(0.0815\) (3 sf) | A1 | Similar scheme for \(P(M - W) < 0\) |
| Total: 4 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(T) = 9 \times 78 + 7 \times 66 \quad (= 1164)$ | B1 | Or $9 \times 78 + 7 \times 66 - 1200$ |
| $\text{Var}(T) = 9 \times 7^2 + 7 \times 5^2 \quad (= 616)$ | B1 | |
| $\dfrac{1200 - \text{'1164'}}{\sqrt{\text{'616'}}} \quad (= 1.450)$ | M1 | $\pm$ Allow without $\sqrt{}$ |
| $P(z < 1.450) = \Phi(1.450)$ | M1 | |
| $= 0.927$ (3 sf) | A1 | Correct tail consistent with their mean |
| **Total: 5** | | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(D) = 66 - 78 \quad (= -12)$ | B1 | Both needed |
| $\text{Var}(D) = 7^2 + 5^2 \quad (= 74)$ | | |
| $\dfrac{0 - (\text{'-12'})}{\sqrt{74}} \quad (= 1.395)$ | M1 | $\pm$ Allow without $\sqrt{}$ |
| $P(D > 0) = 1 - \Phi(\text{'1.395'})$ | M1 | Correct tail consistent with their mean |
| $0.0815$ (3 sf) | A1 | Similar scheme for $P(M - W) < 0$ |
| **Total: 4** | | |
---6 The weights, in kilograms, of men and women have the distributions $\mathrm { N } \left( 78,7 ^ { 2 } \right)$ and $\mathrm { N } \left( 66,5 ^ { 2 } \right)$ respectively.\\
(i) The maximum load that a certain cable car can carry safely is 1200 kg . If 9 randomly chosen men and 7 randomly chosen women enter the cable car, find the probability that the cable car can operate safely.\\
(ii) Find the probability that a randomly chosen woman weighs more than a randomly chosen man. [4]\\
\hfill \mbox{\textit{CAIE S2 Q6 [9]}}