Question 4 - A-Level Maths

AQA Further Paper 3 Statistics 2020 June — Question 4 9 marks

Exam Board	AQA
Module	Further Paper 3 Statistics (Further Paper 3 Statistics)
Year	2020
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Uniform Distribution
Type	Variance of linear transformation
Difficulty	Standard +0.3 This is a straightforward Further Maths statistics question requiring standard variance formula manipulation and basic probability. Part (a) is a guided proof using given formulas, part (b) is trivial probability calculation (3/4), and part (c) asks for a standard assumption. The question is slightly above average difficulty only because it's Further Maths content, but within that context it's routine bookwork with no novel insight required.
Spec	5.02a Discrete probability distributions: general 5.02c Linear coding: effects on mean and variance

4 The discrete random variable $X$ follows a discrete uniform distribution and takes values $1,2,3 , \ldots , n$. The discrete random variable $Y$ is defined by $Y = 2 X$ 4

Using the standard results for $\sum n , \sum n ^ { 2 }$ and $\operatorname { Var } ( a X + b )$, prove that $$\operatorname { Var } ( Y ) = \frac { n ^ { 2 } - 1 } { 3 }$$ 4
A spinning toy can land on one of four values: 2, 4, 6 or 8
Using a discrete uniform distribution, find the probability that the next value the toy lands on is greater than 2 4
State an assumption that is required for the discrete uniform distribution used in part (b) to be valid.

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Question 4:

Part 4(a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$E(X) = \sum_{i=1}^{n}\frac{x}{n} = \frac{1+2+\dots+n}{n}$	M1	Uses formula for $E(X)$
$= \frac{\frac{n}{2}(1+n)}{n} = \frac{n+1}{2}$	A1	Obtains expression using $\sum n$
$E(X^2) = \sum_{i=1}^{n}\frac{x^2}{n} = \frac{1^2+2^2+\dots+n^2}{n}$	M1	Uses formula for $E(X^2)$
$= \frac{\frac{1}{6}n(n+1)(2n+1)}{n} = \frac{(n+1)(2n+1)}{6}$	A1	Obtains expression using $\sum n^2$
$\text{Var}(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$	M1	Uses formula for $\text{Var}(X)$ or uses $\text{Var}(Y) = 2^2\text{Var}(X)$
$= \frac{2n^2+3n+1}{6} - \frac{n^2+2n+1}{4} = \frac{4n^2+6n+2-3n^2-6n-3}{12} = \frac{n^2-1}{12}$	A1	Correct expression for $\text{Var}(X)$
$\text{Var}(Y) = \text{Var}(2X) = 4 \times \frac{n^2-1}{12} = \frac{n^2-1}{3}$	R1	Rigorous algebraic proof using $\text{Var}(Y) = 2^2\text{Var}(X)$ to show $\text{Var}(Y) = \frac{n^2-1}{3}$

Part 4(b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$3 \times \frac{1}{4} = 0.75$	B1	Obtains probability that next value spinner lands on is greater than 2; 0.75 OE

Part 4(c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
The spinner toy is unbiased	E1	Explains that spinner is unbiased or the probability of obtaining each score is equal

## Question 4:

### Part 4(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \sum_{i=1}^{n}\frac{x}{n} = \frac{1+2+\dots+n}{n}$ | M1 | Uses formula for $E(X)$ |
| $= \frac{\frac{n}{2}(1+n)}{n} = \frac{n+1}{2}$ | A1 | Obtains expression using $\sum n$ |
| $E(X^2) = \sum_{i=1}^{n}\frac{x^2}{n} = \frac{1^2+2^2+\dots+n^2}{n}$ | M1 | Uses formula for $E(X^2)$ |
| $= \frac{\frac{1}{6}n(n+1)(2n+1)}{n} = \frac{(n+1)(2n+1)}{6}$ | A1 | Obtains expression using $\sum n^2$ |
| $\text{Var}(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$ | M1 | Uses formula for $\text{Var}(X)$ or uses $\text{Var}(Y) = 2^2\text{Var}(X)$ |
| $= \frac{2n^2+3n+1}{6} - \frac{n^2+2n+1}{4} = \frac{4n^2+6n+2-3n^2-6n-3}{12} = \frac{n^2-1}{12}$ | A1 | Correct expression for $\text{Var}(X)$ |
| $\text{Var}(Y) = \text{Var}(2X) = 4 \times \frac{n^2-1}{12} = \frac{n^2-1}{3}$ | R1 | Rigorous algebraic proof using $\text{Var}(Y) = 2^2\text{Var}(X)$ to show $\text{Var}(Y) = \frac{n^2-1}{3}$ |

### Part 4(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3 \times \frac{1}{4} = 0.75$ | B1 | Obtains probability that next value spinner lands on is greater than 2; 0.75 OE |

### Part 4(c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| The spinner toy is unbiased | E1 | Explains that spinner is unbiased or the probability of obtaining each score is equal |

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Show LaTeX source

4 The discrete random variable $X$ follows a discrete uniform distribution and takes values $1,2,3 , \ldots , n$.

The discrete random variable $Y$ is defined by $Y = 2 X$\\
4
\begin{enumerate}[label=(\alph*)]
\item Using the standard results for $\sum n , \sum n ^ { 2 }$ and $\operatorname { Var } ( a X + b )$, prove that

$$\operatorname { Var } ( Y ) = \frac { n ^ { 2 } - 1 } { 3 }$$

4
\item A spinning toy can land on one of four values: 2, 4, 6 or 8\\
Using a discrete uniform distribution, find the probability that the next value the toy lands on is greater than 2

4
\item State an assumption that is required for the discrete uniform distribution used in part (b) to be valid.
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2020 Q4 [9]}}

This paper (9 questions)

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Q1 1 Q2 1 Q3 4 Q4 9 Q5 7 Q6 8 Q7 8 Q8 6 Q9 6

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(E(X) = \sum_{i=1}^{n}\frac{x}{n} = \frac{1+2+\dots+n}{n}\)	M1	Uses formula for \(E(X)\)
\(= \frac{\frac{n}{2}(1+n)}{n} = \frac{n+1}{2}\)	A1	Obtains expression using \(\sum n\)
\(E(X^2) = \sum_{i=1}^{n}\frac{x^2}{n} = \frac{1^2+2^2+\dots+n^2}{n}\)	M1	Uses formula for \(E(X^2)\)
\(= \frac{\frac{1}{6}n(n+1)(2n+1)}{n} = \frac{(n+1)(2n+1)}{6}\)	A1	Obtains expression using \(\sum n^2\)
\(\text{Var}(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2\)	M1	Uses formula for \(\text{Var}(X)\) or uses \(\text{Var}(Y) = 2^2\text{Var}(X)\)
\(= \frac{2n^2+3n+1}{6} - \frac{n^2+2n+1}{4} = \frac{4n^2+6n+2-3n^2-6n-3}{12} = \frac{n^2-1}{12}\)	A1	Correct expression for \(\text{Var}(X)\)
\(\text{Var}(Y) = \text{Var}(2X) = 4 \times \frac{n^2-1}{12} = \frac{n^2-1}{3}\)	R1	Rigorous algebraic proof using \(\text{Var}(Y) = 2^2\text{Var}(X)\) to show \(\text{Var}(Y) = \frac{n^2-1}{3}\)