| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Standard +0.3 This question requires understanding of linear combinations of normal variables and forming appropriate combinations (3L+4S and L-2S), but the calculations are straightforward applications of standard formulas with no conceptual surprises. Slightly above average difficulty due to the two-part structure and need to correctly set up the second comparison, but well within typical S2 scope. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(L_1+L_2+L_3+S_1+S_2+S_3+S_4) = 3\times5.10+4\times2.51\ [=25.34]\) | B1 | OE (\(E(3L+4S-25.5)=-0.16\)) |
| \(Var(L_1+L_2+L_3+S_1+S_2+S_3+S_4) = 3\times0.0102+4\times0.0036\ [=0.045]\) | B1 | or \(SD=\frac{3\sqrt{2}}{20}=0.2121\) |
| \(\frac{25.5-\text{'25.34'}}{\sqrt{\text{'0.045'}}}\ [=0.754]\) | M1 | No SD/variance mix. Standardising with *their* values (must be from a combination attempt) |
| \(\Phi(\text{'0.754'})\) | M1 | For the correct area consistent with *their* working |
| \(0.775\) (3 sf) | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(L-2S) = 5.10-2\times2.51\ [=0.08]\) | B1 | OE |
| \(Var(L-2S) = 0.0102+2^2\times0.0036\ [=0.0246]\) | B1 | Or \(SD=0.1568\) |
| \(\frac{0-\text{'0.08'}}{\sqrt{\text{'0.0246'}}}\ [=-0.510]\) | M1 | No SD/variance mix. Standardising with *their* values (must be from a combination attempt) |
| \(P(Z>-\text{'0.510'})=\phi(\text{'0.510'})\) | M1 | For the correct area consistent with *their* working |
| \(0.695\) (3 sf) | A1 | |
| 5 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(L_1+L_2+L_3+S_1+S_2+S_3+S_4) = 3\times5.10+4\times2.51\ [=25.34]$ | B1 | OE ($E(3L+4S-25.5)=-0.16$) |
| $Var(L_1+L_2+L_3+S_1+S_2+S_3+S_4) = 3\times0.0102+4\times0.0036\ [=0.045]$ | B1 | or $SD=\frac{3\sqrt{2}}{20}=0.2121$ |
| $\frac{25.5-\text{'25.34'}}{\sqrt{\text{'0.045'}}}\ [=0.754]$ | M1 | No SD/variance mix. Standardising with *their* values (must be from a combination attempt) |
| $\Phi(\text{'0.754'})$ | M1 | For the correct area consistent with *their* working |
| $0.775$ (3 sf) | A1 | |
| | **5** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(L-2S) = 5.10-2\times2.51\ [=0.08]$ | B1 | OE |
| $Var(L-2S) = 0.0102+2^2\times0.0036\ [=0.0246]$ | B1 | Or $SD=0.1568$ |
| $\frac{0-\text{'0.08'}}{\sqrt{\text{'0.0246'}}}\ [=-0.510]$ | M1 | No SD/variance mix. Standardising with *their* values (must be from a combination attempt) |
| $P(Z>-\text{'0.510'})=\phi(\text{'0.510'})$ | M1 | For the correct area consistent with *their* working |
| $0.695$ (3 sf) | A1 | |
| | **5** | |5 The volumes, in litres, of juice in large and small bottles have the distributions $\mathrm { N } ( 5.10,0.0102 )$ and $\mathrm { N } ( 2.51,0.0036 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total volume of juice in 3 randomly chosen large bottles and 4 randomly chosen small bottles is less than 25.5 litres.
\item Find the probability that the volume of juice in a randomly chosen large bottle is at least twice the volume of juice in a randomly chosen small bottle.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q5 [10]}}