| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two or more different variables |
| Difficulty | Standard +0.8 This question requires understanding of linear combinations of independent normal variables (X+Y and X-4Y), applying the variance formula for independent variables, and standardizing to find probabilities. Part (ii) is non-routine as students must recognize that 'X > 4Y' becomes 'X - 4Y > 0' and correctly compute Var(X-4Y) = Var(X) + 16Var(Y). This goes beyond standard textbook exercises and requires problem-solving insight, placing it moderately above average difficulty. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(T) = 234\), \(\text{Var}(T) = 15^2 + 8^2 = 289\) | B1 | |
| \(\frac{200-234}{\sqrt{289}} (= -2.000)\) | M1 | |
| \(\Phi(\text{``}-2.000\text{''}) = 1 - \Phi(\text{``}2.000\text{''})\); \(1 - 0.9772\) | M1 | |
| \(2.28\%\) | A1[4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Require \(P(D>0)\) where \(D = X - 4Y\) | ||
| \(E(D) = 184 - 4 \times 50 = -16\) | B1 | For \(-16\) or \(+16\) or \(\pm(184 - 4\times 50)\) |
| \(\text{Var}(D) = 15^2 + 4^2 \times 8^2 = 1249\) | B1 | For 1249 or \(15^2 + 4^2 \times 8^2\) |
| \(\frac{0-(-16)}{\sqrt{1249}} (= 0.453)\) | M1 | |
| \(1 - \Phi(\text{``}0.453\text{''}) = 1 - 0.6747 = 0.325\) | M1, A1[5] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(T) = 234$, $\text{Var}(T) = 15^2 + 8^2 = 289$ | B1 | |
| $\frac{200-234}{\sqrt{289}} (= -2.000)$ | M1 | |
| $\Phi(\text{``}-2.000\text{''}) = 1 - \Phi(\text{``}2.000\text{''})$; $1 - 0.9772$ | M1 | |
| $2.28\%$ | A1[4] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Require $P(D>0)$ where $D = X - 4Y$ | | |
| $E(D) = 184 - 4 \times 50 = -16$ | B1 | For $-16$ or $+16$ or $\pm(184 - 4\times 50)$ |
| $\text{Var}(D) = 15^2 + 4^2 \times 8^2 = 1249$ | B1 | For 1249 or $15^2 + 4^2 \times 8^2$ |
| $\frac{0-(-16)}{\sqrt{1249}} (= 0.453)$ | M1 | |
| $1 - \Phi(\text{``}0.453\text{''}) = 1 - 0.6747 = 0.325$ | M1, A1[5] | |
---5 Each drink from a coffee machine contains $X \mathrm {~cm} ^ { 3 }$ of coffee and $Y \mathrm {~cm} ^ { 3 }$ of milk, where $X$ and $Y$ are independent variables with $X \sim \mathrm {~N} \left( 184,15 ^ { 2 } \right)$ and $Y \sim \mathrm {~N} \left( 50,8 ^ { 2 } \right)$. If the total volume of the drink is less than $200 \mathrm {~cm} ^ { 3 }$ the customer receives the drink without charge.\\
(i) Find the percentage of drinks which customers receive without charge.\\
(ii) Find the probability that, in a randomly chosen drink, the volume of coffee is more than 4 times the volume of milk.
\hfill \mbox{\textit{CAIE S2 2011 Q5 [9]}}