| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Same variable, two observations |
| Difficulty | Standard +0.8 This S3 question requires understanding of linear combinations of normal variables across multiple parts with increasing complexity. Part (a) needs recognizing D₁-D₂ ~ N(0, 2×1.2²), part (b) requires forming C - 0.8D and finding its distribution, and part (c) involves combining multiple variables (6D + box) - (6C + box). While the techniques are standard for S3, the multi-step reasoning and careful handling of variance rules (especially the 4/5 coefficient in part b) elevate this above routine exercises. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.05c Significance levels: one-tail and two-tail5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Let \(W = D_1 - D_2\) | M1 | |
| \(W \sim N(0, 2.88)\) | A1, A1 | |
| \(P(\ | W\ | > 3) = 2 \times P(W > 3)\) |
| \(= 2 \times P(Z > \frac{3-0}{\sqrt{2.88}})\) | dM1 | |
| \(= 2 \times P(Z > 1.76776....)\) | ||
| \(= 2 \times (1 - 0.9616)\) | ||
| \(= 0.0768\) | awrt 0.077 | A1 |
| (b) Let \(T = 5C - 4D\) or \(4D - 5C\) or \(C - \frac{3}{5}D\) or \(\frac{3}{5}D - C\) | M1 | |
| \(T \sim N(+4, 39.04)\) or \(N(+0.8, 1.5616)\) | A1 A1 | |
| \(P(T < 0) = P(Z < \frac{0-4}{\sqrt{39.04}})\) or \(P(Z < \frac{0-0.8}{\sqrt{1.5616}})\) | M1 | |
| \(= P(Z < -0.64018....)\) | ||
| \(= (1 - 0.7389)\) | ||
| \(= 0.2611\) | awrt 0.261 | A1 |
| (c) Let \(P = D + D + D + D + D + D + B\) | M1 | |
| Let \(Q = C + C + C + C + C + C + B\) | ||
| \(P \sim N(352, 13.64)\) and \(Q \sim N(292, 8.84)\) | A1, A1 | |
| [Let \(R = P - Q\)] | \(R \sim N(\pm 60, 22.48)\) | M1 |
| \(P(R > 50) = P(Z > \frac{50-60}{\sqrt{22.48}})\) | dM1 | |
| \(= P(Z > -2.10....)\) | ||
| \(= 0.9821\) | awrt 0.982 ~ 0.983 | A1 |
**(a)** Let $W = D_1 - D_2$ | M1 |
$W \sim N(0, 2.88)$ | A1, A1 |
$P(\|W\| > 3) = 2 \times P(W > 3)$ | M1 |
$= 2 \times P(Z > \frac{3-0}{\sqrt{2.88}})$ | dM1 |
$= 2 \times P(Z > 1.76776....)$ | |
$= 2 \times (1 - 0.9616)$ | |
$= 0.0768$ | awrt **0.077** | A1 | (6)
**(b)** Let $T = 5C - 4D$ or $4D - 5C$ or $C - \frac{3}{5}D$ or $\frac{3}{5}D - C$ | M1 |
$T \sim N(+4, 39.04)$ or $N(+0.8, 1.5616)$ | A1 A1 |
$P(T < 0) = P(Z < \frac{0-4}{\sqrt{39.04}})$ or $P(Z < \frac{0-0.8}{\sqrt{1.5616}})$ | M1 |
$= P(Z < -0.64018....)$ | |
$= (1 - 0.7389)$ | |
$= 0.2611$ | awrt **0.261** | A1 | (5)
**(c)** Let $P = D + D + D + D + D + D + B$ | M1 |
Let $Q = C + C + C + C + C + C + B$ | |
$P \sim N(352, 13.64)$ and $Q \sim N(292, 8.84)$ | A1, A1 |
[Let $R = P - Q$] | $R \sim N(\pm 60, 22.48)$ | M1 |
$P(R > 50) = P(Z > \frac{50-60}{\sqrt{22.48}})$ | dM1 |
$= P(Z > -2.10....)$ | |
$= 0.9821$ | awrt **0.982 ~ 0.983** | A1 | (6)
**[Total 17]**
**Notes:**
**Award full marks in each part for a correct answer with no incorrect working seen.**
**(a)** 1st M1 for explicitly defining a suitable $W$ and attempt to find the distribution of $W$. May be implied by sight of N(0, 2.88)
1st A1 for normal and mean of 0, 2nd A1 for variance of 2.88. Award M1A1A1 for N(0, 2.88) seen.
3rd M1 for realising need $2 \times P(W > 3)$
3rd dM1 Dep on 1st M1 for standardising with 3, 0 and their s.d. Must lead to P(Z > +ve) (o.e.)
**(b)** 1st M1 for explicitly defining a suitable $T$ but may be implied by sight of one of these normals
1st A1 for normal and correct mean, 2nd A1 for correct variance. Accept awrt 3sf i.e. 39.0, 1.56
2nd M1 for standardising with 0 and their mean and mean and their s.d. Must lead to P(Z < -ve) (o.e.)
**(c)** 1st M1 for explicitly defining a correct $P$ or $Q$. May be implied by a correct distribution for $P$ or $Q$
1st A1 for a correct distribution for $P$ 2nd A1 for a correct distribution for $Q$
2nd M1 for attempting $R$ and obtaining its distribution- ft their $P$ and $Q$ means and variances
3rd dM1 for attempting P(R > 50) and standardising with 50 and their E(R) and their $\sqrt{\text{Var}(R)}$. Dependent on 2nd M1. Must lead to a P(Z > -ve) (o.e.)8. A farmer supplies both duck eggs and chicken eggs. The weights of duck eggs, $D$ grams, and chicken eggs, $C$ grams, are such that
$$D \sim \mathrm {~N} \left( 54,1.2 ^ { 2 } \right) \text { and } C \sim \mathrm {~N} \left( 44,0.8 ^ { 2 } \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the weights of 2 randomly selected duck eggs will differ by more than 3 g .
\item Find the probability that the weight of a randomly selected chicken egg is less than $\frac { 4 } { 5 }$ of the weight of a randomly selected duck egg.
Eggs are packed in boxes which contain either 6 randomly selected duck eggs or 6 randomly selected chicken eggs. The weight of an empty box has distribution $\mathrm { N } \left( 28 , \sqrt { 5 } ^ { 2 } \right)$.
\item Find the probability that a full box of duck eggs weighs at least 50 g more than a full box of chicken eggs.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2013 Q8 [17]}}