| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two or more different variables |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of independent normal distributions with standard formulas. Part (i) requires finding mean and variance of X + 2.5Y, then a single normal probability calculation. Part (ii) similarly requires finding the distribution of X - Y and another probability calculation. Both parts are routine once the student recalls that linear combinations of independent normals are normal, with no conceptual challenges beyond applying the formulas correctly. |
| 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 | Mark | Guidance |
| \(X + 2.5Y \sim N(127,\ 44.25)\) | B1 | B1 for 127; allow at early stage \((57 + 2.5 \times 28)\) |
| B1 | B1 for 44.25 or 6.65; allow at early stage \((13 + 2.5^2 \times 5)\); may be implied by next line | |
| \((\pm)\frac{140 - 127}{\sqrt{44.25}}\) | M1 | For standardising |
| \(= \pm(1.954)\) | M1 | For area consistent with their working |
| \(1 - \Phi(1.954)\) | ||
| \(= 0.0254/0.0253\) (3 s.f.) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X - Y \sim N(29,\ 18)\) | B1 | B1 for 29; give at early stage \((57-28)\) |
| B1 | B1 for 18; give at early stage \((13+5)\); may be implied by next line | |
| \(\frac{20 - 29}{\sqrt{18}} \quad (= -2.121)\) | M1 | For standardising |
| \(1 - \Phi(-2.121) = \Phi(2.121)\) | M1 | For area consistent with their working |
| \(= 0.983\) (3 s.f.) | A1 [5] |
## Question 8:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X + 2.5Y \sim N(127,\ 44.25)$ | B1 | B1 for 127; allow at early stage $(57 + 2.5 \times 28)$ |
| | B1 | B1 for 44.25 or 6.65; allow at early stage $(13 + 2.5^2 \times 5)$; may be implied by next line |
| $(\pm)\frac{140 - 127}{\sqrt{44.25}}$ | M1 | For standardising |
| $= \pm(1.954)$ | M1 | For area consistent with their working |
| $1 - \Phi(1.954)$ | | |
| $= 0.0254/0.0253$ (3 s.f.) | A1 [5] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X - Y \sim N(29,\ 18)$ | B1 | B1 for 29; give at early stage $(57-28)$ |
| | B1 | B1 for 18; give at early stage $(13+5)$; may be implied by next line |
| $\frac{20 - 29}{\sqrt{18}} \quad (= -2.121)$ | M1 | For standardising |
| $1 - \Phi(-2.121) = \Phi(2.121)$ | M1 | For area consistent with their working |
| $= 0.983$ (3 s.f.) | A1 [5] | |8 In an examination, the marks in the theory paper and the marks in the practical paper are denoted by the random variables $X$ and $Y$ respectively, where $X \sim \mathrm {~N} ( 57,13 )$ and $Y \sim \mathrm {~N} ( 28,5 )$. You may assume that each candidate's marks in the two papers are independent. The final score of each candidate is found by calculating $X + 2.5 Y$. A candidate is chosen at random. Without using a continuity correction, find the probability that this candidate\\
(i) has a final score that is greater than 140 ,\\
(ii) obtains at least 20 more marks in the theory paper than in the practical paper.
\hfill \mbox{\textit{CAIE S2 2014 Q8 [10]}}