OCR MEI Further Statistics B AS 2021 November — Question 6 11 marks

Exam BoardOCR MEI
ModuleFurther Statistics B AS (Further Statistics B AS)
Year2021
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeCalculate and compare mean, median, mode
DifficultyStandard +0.3 This is a standard continuous probability distribution question requiring routine integration and application of standard results. Part (a) uses the fundamental property that the pdf integrates to 1, parts (b)-(c) involve straightforward integration to find the CDF, and part (d) requires calculating E(X) and solving F(m)=0.5, then comparingβ€”all textbook techniques with no novel insight required. Slightly above average difficulty due to the multi-part nature and algebraic manipulation needed.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration
6 The probability density function of the continuous random variable \(X\) is given by \(f ( x ) = \begin{cases} 2 ( 1 + a x ) & 0 \leqslant x \leqslant 1 , \\ 0 & \text { otherwise } , \end{cases}\) where \(a\) is a constant.
  1. Show that \(a = - 1\).
  2. Find the cumulative distribution function of \(X\).
  3. Find \(\mathrm { P } ( X < 0.5 )\).
  4. Show that \(\mathrm { E } ( X )\) is greater than the median of \(X\).
Question 6:
AnswerMarks Guidance
6(a) (2(1+0) + 2(1 + a)) = 1
1
a = βˆ’1
2
OR:
β‡’
𝟏𝟏
οΏ½ 𝟐𝟐(𝟏𝟏+𝒂𝒂 𝒂𝒂)𝐝𝐝𝒂𝒂 = 𝟏𝟏
𝟎𝟎
AnswerMarks
𝟐𝟐+𝒂𝒂 = 𝟏𝟏M1
A1
[2]
M1
AnswerMarks Guidance
A12.1
1.1AG OR
1
a = βˆ’1
∫0 2(1+π‘Žπ‘Žπ‘Žπ‘Ž)dπ‘Žπ‘Ž = 1
2 1
[2π‘Žπ‘Ž+π‘Žπ‘Žπ‘Žπ‘Ž ]0 = 1 β‡’
AnswerMarks Guidance
6(b) 𝒂𝒂 = βˆ’πŸπŸ
CDF =
π‘₯π‘₯
=
∫0(2βˆ’2𝑒𝑒)d𝑒𝑒
ο£±02 x<0,
2π‘Žπ‘Žβˆ’ο£΄π‘Žπ‘Ž
F(x)=ο£²2xβˆ’x2 0≀x≀1,

1 x>1.
AnswerMarks
ο£³M1
M1
A1
AnswerMarks
[3]1.1
1.1a
1.1
AnswerMarks Guidance
6(c) P(X < 0.5) = 2 Γ— 0.5 – 0.52
= 0.75M1
A1
AnswerMarks
[2]1.1a
1.1
AnswerMarks Guidance
6(d) E(X) =
1 2
=
∫0(2π‘Žπ‘Žβˆ’2π‘Žπ‘Ž )dπ‘Žπ‘Ž
1
3
m = 0.2923
AnswerMarks
2π‘šπ‘šβˆ’π‘šπ‘š = 0.5M1
A1
M1
A1
AnswerMarks
[4]1.1
1.1
1.1a
AnswerMarks
1.1BC
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Question 6:
6 | (a) | (2(1+0) + 2(1 + a)) = 1
1
a = βˆ’1
2
OR:
β‡’
𝟏𝟏
οΏ½ 𝟐𝟐(𝟏𝟏+𝒂𝒂 𝒂𝒂)𝐝𝐝𝒂𝒂 = 𝟏𝟏
𝟎𝟎
𝟐𝟐+𝒂𝒂 = 𝟏𝟏 | M1
A1
[2]
M1
A1 | 2.1
1.1 | AG | OR
1
a = βˆ’1
∫0 2(1+π‘Žπ‘Žπ‘Žπ‘Ž)dπ‘Žπ‘Ž = 1
2 1
[2π‘Žπ‘Ž+π‘Žπ‘Žπ‘Žπ‘Ž ]0 = 1 β‡’
6 | (b) | 𝒂𝒂 = βˆ’πŸπŸ
CDF =
π‘₯π‘₯
=
∫0(2βˆ’2𝑒𝑒)d𝑒𝑒
ο£±02 x<0,
2π‘Žπ‘Žβˆ’ο£΄π‘Žπ‘Ž
F(x)=ο£²2xβˆ’x2 0≀x≀1,

1 x>1.
ο£³ | M1
M1
A1
[3] | 1.1
1.1a
1.1
6 | (c) | P(X < 0.5) = 2 Γ— 0.5 – 0.52
= 0.75 | M1
A1
[2] | 1.1a
1.1
6 | (d) | E(X) =
1 2
=
∫0(2π‘Žπ‘Žβˆ’2π‘Žπ‘Ž )dπ‘Žπ‘Ž
1
3
m = 0.2923
2π‘šπ‘šβˆ’π‘šπ‘š = 0.5 | M1
A1
M1
A1
[4] | 1.1
1.1
1.1a
1.1 | BC
BC
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
6 The probability density function of the continuous random variable $X$ is given by\\
$f ( x ) = \begin{cases} 2 ( 1 + a x ) & 0 \leqslant x \leqslant 1 , \\ 0 & \text { otherwise } , \end{cases}$\\
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = - 1$.
\item Find the cumulative distribution function of $X$.
\item Find $\mathrm { P } ( X < 0.5 )$.
\item Show that $\mathrm { E } ( X )$ is greater than the median of $X$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2021 Q6 [11]}}
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Q1 9 Q2 10 Q3 12 Q4 9 Q5 9 Q6 11