| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Direct comparison with scalar multiple (different variables) |
| Difficulty | Standard +0.3 This is a standard application of linear combinations of independent normal distributions. Part (a) requires finding the distribution of L+S and a routine probability calculation. Part (b) requires forming L-2S and another standard calculation. Both parts follow textbook procedures with no novel insight required, though the setup in (b) is slightly less obvious than (a), making this slightly easier than average overall. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N(310, 50)\) | B1 | SOI |
| \(\frac{300 - 310}{\sqrt{50}} = -1.414\) | M1 | Standardise using their values |
| \(\Phi(-1.414) = 1 - \phi(1.414)\) | M1 | Area consistent with their values |
| \(= 0.0786\) or \(0.0787\) (3 sf) | A1 | As final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(L - 2S > 0)\) | M1 | OE SOI |
| \(E(X) = 200 - 2 \times 110 = -20\) | B1 | OE seen |
| \(\text{Var} = 30 + 2^2 \times 20 = 110\) | B1 | Seen |
| \(N(-20,\, 110)\); \(\frac{0-(-20)}{\sqrt{110}} = 1.907\) | M1 | Standardising with their values. Mean and variance must come from a combination attempt. |
| \(1 - \Phi(1.907)\) | M1 | Correct area consistent with their working |
| \(= 0.0283\) (3 sf) | A1 | Final answer |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $N(310, 50)$ | B1 | SOI |
| $\frac{300 - 310}{\sqrt{50}} = -1.414$ | M1 | Standardise using their values |
| $\Phi(-1.414) = 1 - \phi(1.414)$ | M1 | Area consistent with their values |
| $= 0.0786$ or $0.0787$ (3 sf) | A1 | As final answer |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(L - 2S > 0)$ | M1 | OE SOI |
| $E(X) = 200 - 2 \times 110 = -20$ | B1 | OE seen |
| $\text{Var} = 30 + 2^2 \times 20 = 110$ | B1 | Seen |
| $N(-20,\, 110)$; $\frac{0-(-20)}{\sqrt{110}} = 1.907$ | M1 | Standardising with their values. Mean and variance must come from a combination attempt. |
| $1 - \Phi(1.907)$ | M1 | Correct area consistent with their working |
| $= 0.0283$ (3 sf) | A1 | Final answer |6 The volumes, in millilitres, of large and small cups of tea are modelled by the distributions $\mathrm { N } ( 200,30 )$ and $\mathrm { N } ( 110,20 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total volume of a randomly chosen large cup of tea and a randomly chosen small cup of tea is less than 300 ml .
\item Find the probability that the volume of a randomly chosen large cup of tea is more than twice the volume of a randomly chosen small cup of tea.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q6 [10]}}