| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Different variables, one observation each |
| Difficulty | Challenging +1.2 This question tests standard variance properties of sums and differences of correlated normal variables. Part (a)(i) is algebraic manipulation using Var(X₁+X₂) = Var(X₁) + Var(X₂) + 2Cov(X₁,X₂). Part (a)(ii) and (b) are routine normal probability calculations once the variance is found. While it requires understanding covariance (an S3 topic), the execution is mechanical with no novel insight needed—slightly above average difficulty due to the covariance concept but still a standard textbook exercise. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(Y) = \text{Var}(X_1) + \text{Var}(X_2) + 2\text{Cov}(X_1, X_2)\) | M1 | Correct variance formula for sum |
| \(140^2 = 120^2 + 120^2 + 2\text{Cov}(X_1, X_2)\) | A1 | Correct substitution |
| \(19600 = 14400 + 14400 + 2\text{Cov}(X_1, X_2)\) | ||
| \(\text{Cov}(X_1, X_2) = \frac{19600 - 28800}{2} = -4600\) | A1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X_1 - X_2) = 1000 - 1000 = 0\) | B1 | Correct mean |
| \(\text{Var}(X_1 - X_2) = \text{Var}(X_1) + \text{Var}(X_2) - 2\text{Cov}(X_1,X_2)\) | M1 | Correct formula |
| \(= 14400 + 14400 - 2(-4600) = 38000\) | A1 | Correct variance |
| \(P(X_1 - X_2 > 250) = P\!\left(Z > \frac{250}{\sqrt{38000}}\right)\) | M1 | Standardising correctly |
| \(= P(Z > 1.2825...)\) | ||
| \(= 1 - \Phi(1.283) = 1 - 0.9002 = 0.0998\) | A1 | Accept 0.0997–0.100 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(W = Y + B\) where \(B \sim N(500, 40^2)\) | M1 | Correct combination |
| \(E(W) = 2000 + 500 = 2500\) | B1 | Correct mean |
| \(\text{Var}(W) = 140^2 + 40^2 = 19600 + 1600 = 21200\) | M1 | Correct variance (independent so no covariance term) |
| \(P(W < 2750) = P\!\left(Z < \frac{2750 - 2500}{\sqrt{21200}}\right)\) | M1 | Standardising correctly |
| \(= P(Z < 1.7174...)\) | ||
| \(= \Phi(1.717) = 0.9570\) | A1 | Accept 0.957–0.958 |
# Question 6:
## Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(Y) = \text{Var}(X_1) + \text{Var}(X_2) + 2\text{Cov}(X_1, X_2)$ | M1 | Correct variance formula for sum |
| $140^2 = 120^2 + 120^2 + 2\text{Cov}(X_1, X_2)$ | A1 | Correct substitution |
| $19600 = 14400 + 14400 + 2\text{Cov}(X_1, X_2)$ | | |
| $\text{Cov}(X_1, X_2) = \frac{19600 - 28800}{2} = -4600$ | A1 | Correctly shown |
## Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X_1 - X_2) = 1000 - 1000 = 0$ | B1 | Correct mean |
| $\text{Var}(X_1 - X_2) = \text{Var}(X_1) + \text{Var}(X_2) - 2\text{Cov}(X_1,X_2)$ | M1 | Correct formula |
| $= 14400 + 14400 - 2(-4600) = 38000$ | A1 | Correct variance |
| $P(X_1 - X_2 > 250) = P\!\left(Z > \frac{250}{\sqrt{38000}}\right)$ | M1 | Standardising correctly |
| $= P(Z > 1.2825...)$ | | |
| $= 1 - \Phi(1.283) = 1 - 0.9002 = 0.0998$ | A1 | Accept 0.0997–0.100 |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $W = Y + B$ where $B \sim N(500, 40^2)$ | M1 | Correct combination |
| $E(W) = 2000 + 500 = 2500$ | B1 | Correct mean |
| $\text{Var}(W) = 140^2 + 40^2 = 19600 + 1600 = 21200$ | M1 | Correct variance (independent so no covariance term) |
| $P(W < 2750) = P\!\left(Z < \frac{2750 - 2500}{\sqrt{21200}}\right)$ | M1 | Standardising correctly |
| $= P(Z < 1.7174...)$ | | |
| $= \Phi(1.717) = 0.9570$ | A1 | Accept 0.957–0.958 |6 The weight, $X$ grams, of a dressed pheasant may be modelled by a normal random variable with a mean of 1000 and a standard deviation of 120 .
Pairs of dressed pheasants are selected for packing into boxes. The total weight of a pair, $Y = X _ { 1 } + X _ { 2 }$ grams, may be modelled by a normal distribution with a mean of 2000 and a standard deviation of 140 .
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\operatorname { Cov } \left( X _ { 1 } , X _ { 2 } \right) = - 4600$.
\item Given that $X _ { 1 } - X _ { 2 }$ may be assumed to be normally distributed, determine the probability that the difference between the weights of a selected pair of dressed pheasants exceeds 250 grams.
\end{enumerate}\item The weight of a box is independent of the total weight of a pair of dressed pheasants, and is normally distributed with a mean of 500 grams and a standard deviation of 40 grams.
Determine the probability that a box containing a pair of dressed pheasants weighs less than 2750 grams.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{fa3bf9d6-f064-4214-acff-d8b88c33a81e-16_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2011 Q6 [13]}}