AQA Further AS Paper 2 Statistics 2020 June — Question 6 8 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Statistics (Further AS Paper 2 Statistics)
Year2020
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeCalculate probability P(X in interval)
DifficultyStandard +0.3 This is a straightforward Further Maths statistics question requiring standard integration of a polynomial pdf. Part (a) is routine integration, part (b) involves verifying a given median (reducing computational burden), and part (c) requires computing E(X) by integration. While the polynomial is cubic, the techniques are mechanical and well-practiced. Slightly above average difficulty due to the algebraic manipulation required, but no conceptual challenges or novel problem-solving.
Spec5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles
The continuous random variable \(X\) has probability density function $$f(x) = \begin{cases} \frac{4}{45}(x^3 - 10x^2 + 29x - 20) & 1 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$$
  1. Find P\((X < 2)\) [2 marks]
  2. Verify that the median of \(X\) is 2.3, correct to two significant figures. [4 marks]
  3. Find the mean of \(X\). [2 marks]
Question 6:
AnswerMarks Guidance
6(a)Uses an integral of f(x) with
one limit of 21.1a M1
P(X < 2) = ∫ (x3 −10x2 +29x−20) dx
45
1
2
4 x4 10x3 29x2 
=  − + −20x
45 4 3 2 
1
47
=
135
47
Obtains P(X < 2) = or
135
AnswerMarks Guidance
AWRT 0.3481.1b A1
AnswerMarks
6(b)Integrates
m 4
∫ (x3 −10x2 +29x−20) dx
45
1
AnswerMarks Guidance
and compares with 0.51.1a M1
∫ (x3 −10x2 +29x−20) dx=0.5
45
1
m
4 x4 10x3 29x2 
− + −20x =0.5
45 4 3 2 
1
1 8 58 16 103
m4 − m3 + m2 − m+ =0.5
45 27 45 9 135
1 8 58 16 71
m4 − m3 + m2 − m+ =0
45 27 45 9 270
m = 2.282817
Therefore, the median of X is 2.3 to two
significant figures.
Obtains correct quartic
equation in terms of the
median.
AnswerMarks Guidance
PI by sight of 2.281.1b A1
Obtains equation such that
one side is equal to zero.
AnswerMarks Guidance
PI by sight of 2.281.1b A1
Concludes correctly that the
median of X is 2.3 to two
significant figures because
either the median is equal to
AWRT 2.28 or that there is a
sign change OE when a
value of m such that 2.25 ≤ m
< 2.3 and a value of m such
that 2.3 < m ≤ 2.35 are
AnswerMarks Guidance
substituted into the equation.2.1 R1
AnswerMarks
6(c)Uses a correct integral of the
form ∫xf(x) dx with any
AnswerMarks Guidance
limits.1.1a M1
E(X) = ∫ x(x3 −10x2 +29x−20) dx
45
1
4
4 x5 10x4 29x3 20x2
=  − + − 
45 5 4 3 2 
1
= 2.32
AnswerMarks Guidance
Obtains E(X) = 2.32 OE1.1b A1
Total8
QMarking Instructions AO 
Question 6:
--- 6(a) ---
6(a) | Uses an integral of f(x) with
one limit of 2 | 1.1a | M1 | 2 4
P(X < 2) = ∫ (x3 −10x2 +29x−20) dx
45
1
2
4 x4 10x3 29x2 
=  − + −20x
45 4 3 2 
1
47
=
135
47
Obtains P(X < 2) = or
135
AWRT 0.348 | 1.1b | A1
--- 6(b) ---
6(b) | Integrates
m 4
∫ (x3 −10x2 +29x−20) dx
45
1
and compares with 0.5 | 1.1a | M1 | m 4
∫ (x3 −10x2 +29x−20) dx=0.5
45
1
m
4 x4 10x3 29x2 
− + −20x =0.5

45 4 3 2 
1
1 8 58 16 103
m4 − m3 + m2 − m+ =0.5
45 27 45 9 135
1 8 58 16 71
m4 − m3 + m2 − m+ =0
45 27 45 9 270
m = 2.282817
Therefore, the median of X is 2.3 to two
significant figures.
Obtains correct quartic
equation in terms of the
median.
PI by sight of 2.28 | 1.1b | A1
Obtains equation such that
one side is equal to zero.
PI by sight of 2.28 | 1.1b | A1
Concludes correctly that the
median of X is 2.3 to two
significant figures because
either the median is equal to
AWRT 2.28 or that there is a
sign change OE when a
value of m such that 2.25 ≤ m
< 2.3 and a value of m such
that 2.3 < m ≤ 2.35 are
substituted into the equation. | 2.1 | R1
--- 6(c) ---
6(c) | Uses a correct integral of the
form ∫xf(x) dx with any
limits. | 1.1a | M1 | 4 4
E(X) = ∫ x(x3 −10x2 +29x−20) dx
45
1
4
4 x5 10x4 29x3 20x2
=  − + − 
45 5 4 3 2 
1
= 2.32
Obtains E(X) = 2.32 OE | 1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
The continuous random variable $X$ has probability density function

$$f(x) = \begin{cases} 
\frac{4}{45}(x^3 - 10x^2 + 29x - 20) & 1 \leq x \leq 4 \\
0 & \text{otherwise}
\end{cases}$$

\begin{enumerate}[label=(\alph*)]
\item Find P$(X < 2)$
[2 marks]

\item Verify that the median of $X$ is 2.3, correct to two significant figures.
[4 marks]

\item Find the mean of $X$.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2020 Q6 [8]}}
This paper (8 questions)
View full paper
Q1 1 Q2 1 Q3 5 Q4 3 Q5 7 Q6 8 Q7 6 Q8 9