| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Same variable, two observations |
| Difficulty | Standard +0.8 Part (i) is a standard application of sums of normal variables requiring only the formula for sum of n independent normals. Part (ii) is more sophisticated, requiring students to recognize they need the distribution of X - 1.1Y (a linear combination with different coefficients), calculate its variance correctly using Var(aX + bY) = a²Var(X) + b²Var(Y), and interpret the problem correctly. This goes beyond routine exercises and requires genuine problem-solving insight. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T \sim N(6 \times 2.4,\ 6 \times 0.3^2)\) \((= N(14.4, 0.54))\) | M1 | seen or implied |
| \(\frac{16 - `14.4`}{\sqrt{`0.54`}}\ (= 2.177)\) | M1 | ft their E and Var; allow without \(\sqrt{}\); Accept alternative method \(N(2.4, (0.3^2)/6)\) |
| \(1 -\ (`2.177`)\) | M1 | correct area consistent with their working |
| \(= 0.0147\) (3 sf) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(D = X_1 - 1.1X_2\); \(E(D) = -0.24\) | B1 | |
| \(\text{Var}(D) = 0.3^2 + 1.1^2 \times 0.3^2\ (= 0.1989)\) | M1 | |
| \(\frac{0 - (-0.24)}{\sqrt{`0.1989`}}\ (= 0.538)\) | M1 | ft their E and Var; allow without \(\sqrt{}\) |
| \((`0.538`)\) | M1 | correct area consistent with their working |
| \(= 0.705\) (3 sf) | A1 [5] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T \sim N(6 \times 2.4,\ 6 \times 0.3^2)$ $(= N(14.4, 0.54))$ | M1 | seen or implied |
| $\frac{16 - `14.4`}{\sqrt{`0.54`}}\ (= 2.177)$ | M1 | ft their E and Var; allow without $\sqrt{}$; Accept alternative method $N(2.4, (0.3^2)/6)$ |
| $1 -\ (`2.177`)$ | M1 | correct area consistent with their working |
| $= 0.0147$ (3 sf) | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D = X_1 - 1.1X_2$; $E(D) = -0.24$ | B1 | |
| $\text{Var}(D) = 0.3^2 + 1.1^2 \times 0.3^2\ (= 0.1989)$ | M1 | |
| $\frac{0 - (-0.24)}{\sqrt{`0.1989`}}\ (= 0.538)$ | M1 | ft their E and Var; allow without $\sqrt{}$ |
| $(`0.538`)$ | M1 | correct area consistent with their working |
| $= 0.705$ (3 sf) | A1 [5] | |
---5 The thickness of books in a large library is normally distributed with mean 2.4 cm and standard deviation 0.3 cm .\\
(i) Find the probability that the total thickness of 6 randomly chosen books is more than 16 cm .\\
(ii) Find the probability that the thickness of a book chosen at random is less than 1.1 times the thickness of a second book chosen at random.
\hfill \mbox{\textit{CAIE S2 2016 Q5 [9]}}