| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal distributions with clear structure. Part (a) requires finding the distribution of 3L + 6S and calculating a single probability; part (b) requires finding the distribution of L - 2S and another probability calculation. Both parts use standard formulas for means and variances of linear combinations with no conceptual challenges or novel problem-solving required—slightly easier than average due to the routine nature of the calculations. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | [ ] | |
| E(T)=3×55+6×27 =327 | B1 | OE. Accept unsimplified. |
| Var(T) = 3×32 + 6×2.52 [= 64.5] | B1 | Accept unsimplified. |
| Answer | Marks | Guidance |
|---|---|---|
| '64.5' | M1 | Must have √ |
| P(z < '1.619') = Φ('1.619') | M1 | Correct probability area consistent with their |
| Answer | Marks |
|---|---|
| 0.947 (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | E(L–S –S ) = 55 – 2 × 27 [=1] | |
| 1 2 | B1 | OE e.g. E(S +S – L)= –1. Accept unsimplified. |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 | B1 | Accept unsimplified. |
| Answer | Marks | Guidance |
|---|---|---|
| '21.5' | M1 | Standardising with their values. Must come from a |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 | M1 | Correct probability area consistent with their |
| Answer | Marks |
|---|---|
| 0.586 or 0.585 (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(a) ---
7(a) | [ ]
E(T)=3×55+6×27 =327 | B1 | OE. Accept unsimplified.
Var(T) = 3×32 + 6×2.52 [= 64.5] | B1 | Accept unsimplified.
340−'327'
[= 1.619]
'64.5' | M1 | Must have √
P(z < '1.619') = Φ('1.619') | M1 | Correct probability area consistent with their
working.
0.947 (3 sf) | A1
5
--- 7(b) ---
7(b) | E(L–S –S ) = 55 – 2 × 27 [=1]
1 2 | B1 | OE e.g. E(S +S – L)= –1. Accept unsimplified.
1 2
Var(L– S –S ) = 32 + 2 × 2.52 [= 21.5]
1 2 | B1 | Accept unsimplified.
0−'1'
[= –0.216]
'21.5' | M1 | Standardising with their values. Must come from a
combination attempt.
P(L–S –S > 0) = Φ(‘0.216’)
1 2 | M1 | Correct probability area consistent with their
working.
0.586 or 0.585 (3 sf) | A1
5
Question | Answer | Marks | GuidanceThe masses, in kilograms, of large and small sacks of flour have the distributions $\text{N}(55, 3^2)$ and $\text{N}(27, 2.5^2)$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Some sacks are loaded onto a boat. The maximum load of flour that the boat can carry safely is 340 kg.
Find the probability that the boat can carry safely 3 randomly chosen large sacks of flour and 6 randomly chosen small sacks of flour. [5]
\item Find the probability that the mass of a randomly chosen large sack of flour is greater than the total mass of two randomly chosen small sacks of flour. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q7 [10]}}