| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Distribution of linear combination |
| Difficulty | Standard +0.8 This is a multi-part Further Maths Statistics question requiring understanding of linear combinations of normal distributions, conditional probability with discrete cases, and solving simultaneous equations from distribution parameters. Part (c) particularly requires algebraic manipulation of mean and variance formulas with two unknowns, which is more demanding than routine application of formulas. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((A+R) \sim N(300,\, 12^2+10^2)\) | M1 | Setting up either model for the weights of the two fruit |
| \((A_1+A_2) \sim N(320,\, 2\times12^2)\) | A1, A1 | Correct distribution for 1 apple 1 orange; correct distribution for 2 apples |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(\text{both apples}) = \frac{4}{5}\times\frac{3}{4} = \frac{3}{5}\); \(P(\text{one apple and one orange}) = \frac{2}{5}\) | M1 | Finding probability for each possible outcome |
| \(\frac{3}{5}{'}P(A_1+A_2>310){'} + \frac{2}{5}{'}P(A+R>310{)}\) | M1 | Fully correct method for finding the required probability |
| \(= 0.5377...\) awrt 0.538 | A1 | awrt 0.538 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left[W = \sum_1^m A - n\times R\right]\); \(W \sim N(160m-140n,\; m\times12^2 + n^2\times10^2)\) | M1, A1 | Setting up model for \(W\); correct distribution |
| \(160m - 140n = (1100.08+1499.92)\div 2 = 1300\) | M1 | Using given interval to set up equation for mean |
| \(2\times1.96\times\sqrt{m\times12^2+n^2\times10^2} = (1499.92-1100.08)\); \(\left[\sqrt{m\times144+n^2\times100} = 102\right]\) | B1, M1 | 1.96; using given interval to set up equation for variance |
| \(m = \frac{1300+140n}{160} \rightarrow \sqrt{\left(\frac{1300+140n}{160}\right)\times144 + n^2\times100} = 102\); \(100n^2 + 126n - 9234 = 0\) | dM1 | Solving simultaneously leading to a 3TQ (dep on previous M mark) |
| \(n = 9\) (\(n=-10.26\) reject) | A1 | \(n=9\) only |
| \(m = 16\) | A1 | \(m=16\) only |
# Question 7:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $(A+R) \sim N(300,\, 12^2+10^2)$ | M1 | Setting up either model for the weights of the two fruit |
| $(A_1+A_2) \sim N(320,\, 2\times12^2)$ | A1, A1 | Correct distribution for 1 apple 1 orange; correct distribution for 2 apples |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(\text{both apples}) = \frac{4}{5}\times\frac{3}{4} = \frac{3}{5}$; $P(\text{one apple and one orange}) = \frac{2}{5}$ | M1 | Finding probability for each possible outcome |
| $\frac{3}{5}{'}P(A_1+A_2>310){'} + \frac{2}{5}{'}P(A+R>310{)}$ | M1 | Fully correct method for finding the required probability |
| $= 0.5377...$ awrt **0.538** | A1 | awrt 0.538 |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\left[W = \sum_1^m A - n\times R\right]$; $W \sim N(160m-140n,\; m\times12^2 + n^2\times10^2)$ | M1, A1 | Setting up model for $W$; correct distribution |
| $160m - 140n = (1100.08+1499.92)\div 2 = 1300$ | M1 | Using given interval to set up equation for mean |
| $2\times1.96\times\sqrt{m\times12^2+n^2\times10^2} = (1499.92-1100.08)$; $\left[\sqrt{m\times144+n^2\times100} = 102\right]$ | B1, M1 | 1.96; using given interval to set up equation for variance |
| $m = \frac{1300+140n}{160} \rightarrow \sqrt{\left(\frac{1300+140n}{160}\right)\times144 + n^2\times100} = 102$; $100n^2 + 126n - 9234 = 0$ | dM1 | Solving simultaneously leading to a 3TQ (dep on previous M mark) |
| $n = 9$ ($n=-10.26$ reject) | A1 | $n=9$ only |
| $m = 16$ | A1 | $m=16$ only |\begin{enumerate}
\item The weights of a particular type of apple, $A$ grams, and a particular type of orange, $R$ grams, each follow independent normal distributions.
\end{enumerate}
$$A \sim \mathrm {~N} \left( 160,12 ^ { 2 } \right) \quad R \sim \mathrm {~N} \left( 140,10 ^ { 2 } \right)$$
(a) Find the distribution of\\
(i) $A + R$\\
(ii) the total weight of 2 randomly selected apples.
A box contains 4 apples and 1 orange only. Jesse selects 2 pieces of fruit at random from the box.\\
(b) Find the probability that the total weight of the 2 pieces of fruit exceeds 310 grams.
From a large number of apples and oranges, Celeste selects $m$ apples and 1 orange at random. The random variable $W$ is given by
$$W = \left( \sum _ { i = 1 } ^ { m } A _ { i } \right) - n \times R$$
where $n$ is a positive integer.\\
Given that the middle $95 \%$ of the distribution of $W$ lies between 1100.08 and 1499.92 grams, (c) find the value of $m$ and the value of $n$
\hfill \mbox{\textit{Edexcel FS2 2021 Q7 [14]}}