| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Different variables, one observation each |
| Difficulty | Standard +0.8 This S3 question requires forming the distribution of S - 2C (non-trivial linear combination with coefficient ≠1), then a probability calculation, followed by a sum of 100 normals. Both parts require careful setup of variance calculations (Var(S-2C) = 0.64 + 4(0.16)) and understanding independence. More conceptually demanding than routine single-variable normal problems, but follows standard S3 techniques without requiring novel insight. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(S > 2C) = P(S - 2C > 0)\) | B1 | |
| \(E[S - 2C] = 4.9 - 2 \times 2.5 = -0.1\) | M1A1 | |
| \(\text{Var}(S - 2C) = 0.64 + 4 \times 0.16 = 1.28\) | M1, M1 | |
| \(P(S - 2C > 0), = P\left(Z > \frac{0-(-0.1)}{\sqrt{1.28}}\right)\) | ||
| \(= P(Z > 0.08838...)\) | A1 | (6) |
| \(= 0.4641\) (tables), or \(0.4648...\) (calculator) accept awrt 0.464 or 0.465 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(T = S_1 + S_2 + ... + S_{100}\) | M1A1 | |
| \(E[T] = 100 \times 4.9 = 490\) | A1 | |
| \(\text{Var}(T) = 100 \times 0.64 = 64\) | M1 | |
| \(P(T < 500) = P\left(Z < \frac{500-490}{\sqrt{64}}\right)\) | ||
| \(= P(Z < 1.25)\) | A1 | (5) |
| \(= 0.8944\) | (11 marks) |
**3(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(S > 2C) = P(S - 2C > 0)$ | B1 | |
| $E[S - 2C] = 4.9 - 2 \times 2.5 = -0.1$ | M1A1 | |
| $\text{Var}(S - 2C) = 0.64 + 4 \times 0.16 = 1.28$ | M1, M1 | |
| $P(S - 2C > 0), = P\left(Z > \frac{0-(-0.1)}{\sqrt{1.28}}\right)$ | | |
| $= P(Z > 0.08838...)$ | A1 | (6) |
| $= 0.4641$ (tables), or $0.4648...$ (calculator) accept awrt 0.464 or 0.465 | | |
**3(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $T = S_1 + S_2 + ... + S_{100}$ | M1A1 | |
| $E[T] = 100 \times 4.9 = 490$ | A1 | |
| $\text{Var}(T) = 100 \times 0.64 = 64$ | M1 | |
| $P(T < 500) = P\left(Z < \frac{500-490}{\sqrt{64}}\right)$ | | |
| $= P(Z < 1.25)$ | A1 | (5) |
| $= 0.8944$ | | (11 marks) |
**Guidance Notes:**
- 1st M1 for $...±4\text{Var}(C)$
- 2nd M1 for $P(S - 2C > 0)$
- 3rd M1 ft their expectation and variance but not if Var($S - 2C$) is negative. (Should lead to $P(Z > +ve)$)
- 1st M1 for attempt to find mean or variance of total
- 1st A1 either correct
- 2nd A1 both correct
- 2nd M1 for standardising using 500, their mean and their sd leading to $P(Z < +ve)$ o.e.
---\begin{enumerate}
\item A company produces two types of milk powder, 'Semi-Skimmed' and 'Full Cream'. In tests, each type of milk powder is used to make a large number of cups of coffee. The mass, $S$ grams, of 'Semi-Skimmed' milk powder used in one cup of coffee is modelled by $S \sim \mathrm {~N} \left( 4.9,0.8 ^ { 2 } \right)$. The mass, $C$ grams, of 'Full Cream' milk powder used in one cup of coffee is modelled by $C \sim \mathrm {~N} \left( 2.5,0.4 ^ { 2 } \right)$\\
(a) Two cups of coffee, one with each type of milk powder, are to be selected at random. Find the probability that the mass of 'Semi-Skimmed' milk powder used will be at least double that of the 'Full Cream' milk powder used.\\
(b) 'Semi-Skimmed' milk powder is sold in 500 g packs. Find the probability that one pack will be sufficient for 100 cups of coffee.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2014 Q3 [11]}}