| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | All components random including container |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal distributions. Part (a) requires summing three independent normals (standard technique), while part (b) involves forming S - 1.4R ~ N(μ, σ²) and finding a probability. Both are direct applications of taught methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T \sim N(515, 74)\) | B1 | B1 for N(515,..) give at early stage |
| B1 | B1 for Var \(= 45 + 25 + 4 = 74\) give at early stage | |
| \(\dfrac{500 - \text{'515'}}{\sqrt{\text{'74'}}}\) \([= -1.744]\) | M1 | Standardise with *their* values; no standard deviation/variance mix; need combination for variance; allow continuity correction |
| \(\Phi(\text{'1.744'})\) | M1 | Area consistent with *their* working |
| 0.959 or 0.96[0] (3 s.f.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(S - 1.4R) = 300 - 1.4 \times 200 = 20\) | B1 | Give at early stage |
| \(\text{Var}(S - 1.4R) = 45 + 1.4^2 \times 25 = 94\) | B1 | Give at early stage; SC: if B0B0 awarded allow SC B1 for 14 and 105.84 |
| \(\dfrac{0 - (20)}{\sqrt{\text{'94'}}}\) \([= -2.063]\) | M1 | Standardise with *their* values; no standard deviation/variance mix; need combination for variance |
| \(1 - \Phi(\text{'2.063'})\) | M1 | Area consistent with *their* working |
| 0.0196 (3 s.f.) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T \sim N(515, 74)$ | B1 | B1 for N(515,..) give at early stage |
| | B1 | B1 for Var $= 45 + 25 + 4 = 74$ give at early stage |
| $\dfrac{500 - \text{'515'}}{\sqrt{\text{'74'}}}$ $[= -1.744]$ | M1 | Standardise with *their* values; no standard deviation/variance mix; need combination for variance; allow continuity correction |
| $\Phi(\text{'1.744'})$ | M1 | Area consistent with *their* working |
| 0.959 or 0.96[0] (3 s.f.) | A1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(S - 1.4R) = 300 - 1.4 \times 200 = 20$ | B1 | Give at early stage |
| $\text{Var}(S - 1.4R) = 45 + 1.4^2 \times 25 = 94$ | B1 | Give at early stage; SC: if B0B0 awarded allow SC B1 for 14 and 105.84 |
| $\dfrac{0 - (20)}{\sqrt{\text{'94'}}}$ $[= -2.063]$ | M1 | Standardise with *their* values; no standard deviation/variance mix; need combination for variance |
| $1 - \Phi(\text{'2.063'})$ | M1 | Area consistent with *their* working |
| 0.0196 (3 s.f.) | A1 | |
---4 Each box of Seeds \& Raisins contains $S$ grams of seeds and $R$ grams of raisins. The weight of a box, when empty, is $B$ grams. $S , R$ and $B$ are independent random variables, where $S \sim \mathrm {~N} ( 300,45 )$, $R \sim \mathrm {~N} ( 200,25 )$ and $\mathrm { B } \sim \mathrm { N } ( 15,4 )$. A full box of Seeds \& Raisins is chosen at random.\\[0pt]
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total weight of the box and its contents is more than 500 grams. [5]
\item Find the probability that the weight of seeds in the box is less than 1.4 times the weight of raisins in the box.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q4 [10]}}