OCR Further Statistics 2019 June — Question 9 14 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2019
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeCDF of transformed variable
DifficultyStandard +0.8 This is a Further Maths statistics question requiring multiple sophisticated techniques: finding the CDF of a transformed variable (standard but requires careful reasoning about inequalities), deriving a moment generating function with domain restrictions (requires integration and convergence analysis), and connecting exponential and Poisson distributions conceptually. The final part requires recognizing the relationship P(T>θ) = e^(-λ) which is non-trivial. More demanding than typical A-level questions but within reach for well-prepared FM students.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03d E(g(X)): general expectation formula5.03e Find cdf: by integration
9 The continuous random variable \(T\) has cumulative distribution function \(F ( t ) = \begin{cases} 0 & t < 0 , \\ 1 - \mathrm { e } ^ { - 0.25 t } & t \geqslant 0 . \end{cases}\)
  1. Find the cumulative distribution function of \(2 T\).
  2. Show that, for constant \(k , \mathrm { E } \left( \mathrm { e } ^ { k t } \right) = \frac { 1 } { 1 - 4 k }\). You should state with a reason the range of values of \(k\) for which this result is valid.
  3. \(\quad T\) is the time before a certain event occurs. Show that the probability that no event occurs between time \(T = 0\) and time \(T = \theta\) is the same as the probability that the value of a random variable with the distribution \(\operatorname { Po } ( \lambda )\) is 0 , for a certain value of \(\lambda\). You should state this value of \(\lambda\) in terms of \(\theta\). \section*{END OF QUESTION PAPER}
Question 9:
AnswerMarks Guidance
9(a) Let H(x) be the CDF of 2T. Then
H(x) = P(X  x) = P(2T  x)
= P(T  ½x) = F(½x)
AnswerMarks
= 1e0.125x [for x  0, and 0 for x < 0]M1
M1
A1
AnswerMarks
[3]3.1a
1.1a
AnswerMarks
1.1Convert to P(2T  x)
Rearrange to get P(T  f(x))
Any letter. Correct answer only,
AnswerMarks
ignore other rangesAlternatively:
g(T) = 2T, F(g--1(x)): M2
Needn’t be simplified
AnswerMarks
(b)Due to the error on the paper, all candidates get 7 marks for Q9(b). Annotate each answer with SEEN and enter 7 marks in RM. The only instance
where full marks would not be awarded is where a candidate has not attempted any question. This would then need to be a 0.
PDF is f(x) = 0.25e–0.25t
E(ekt) =   0.25e kt e 0.25t dt
0
N
 N 0.25e (0.25k)t dt=   0.25e (0.25k)t  
0  0.25k 
0
(0.25k)N
0.25e 0.25
= 
0.25k 0.25k
The first term will only converge for k < 0.25
(0.25k)N
0.25e
Then lim 0
N 0.25k
  0.25e (0.25k)t dt
0
 0.25e (0.25k)N 0.25 
 lim   
N 0.25k 0.25k
0.25 1
= or AG
AnswerMarks
0.25k 14kM1
M1
M1
A1
B1
B1
A1
AnswerMarks
[7]2.1
1.1a
2.1
1.1
2.4
2.1
AnswerMarks
2.2aStated or implied
Attempt ektf(t)dt, correct limits
Method for integration
Correct indefinite integral with finite
upper limit
Consider range of k for which the
result is valid
Correctly obtain given answer
AnswerMarks
(c)P(no event between 0 and ) = P(T > )
= e–0.25
P(0) from Po() = e–
AnswerMarks
Hence same expression, with  = 0.25.M1
A1
B1
A1
AnswerMarks
[4]2.1
1.1
1.1
AnswerMarks
2.2aCorrect method for probability
Correct formula
Simplified, any 
Correctly justify required result,
AnswerMarks
with  = 0.25 oe stated explicitly1 – e–0.25: M1A0
i.e. neither 0! nor e0 left in
Need to say “same” oe
PPMMTT
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
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programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
Question 9:
9 | (a) | Let H(x) be the CDF of 2T. Then
H(x) = P(X  x) = P(2T  x)
= P(T  ½x) = F(½x)
= 1e0.125x [for x  0, and 0 for x < 0] | M1
M1
A1
[3] | 3.1a
1.1a
1.1 | Convert to P(2T  x)
Rearrange to get P(T  f(x))
Any letter. Correct answer only,
ignore other ranges | Alternatively:
g(T) = 2T, F(g--1(x)): M2
Needn’t be simplified
(b) | Due to the error on the paper, all candidates get 7 marks for Q9(b). Annotate each answer with SEEN and enter 7 marks in RM. The only instance
where full marks would not be awarded is where a candidate has not attempted any question. This would then need to be a 0.
PDF is f(x) = 0.25e–0.25t
E(ekt) =   0.25e kt e 0.25t dt
0
N
 N 0.25e (0.25k)t dt=   0.25e (0.25k)t  
0  0.25k 
0
(0.25k)N
0.25e 0.25
= 
0.25k 0.25k
The first term will only converge for k < 0.25
(0.25k)N
0.25e
Then lim 0
N 0.25k
  0.25e (0.25k)t dt
0
 0.25e (0.25k)N 0.25 
 lim   
N 0.25k 0.25k
0.25 1
= or AG
0.25k 14k | M1
M1
M1
A1
B1
B1
A1
[7] | 2.1
1.1a
2.1
1.1
2.4
2.1
2.2a | Stated or implied
Attempt ektf(t)dt, correct limits
Method for integration
Correct indefinite integral with finite
upper limit
Consider range of k for which the
result is valid
Correctly obtain given answer
(c) | P(no event between 0 and ) = P(T > )
= e–0.25
P(0) from Po() = e–
Hence same expression, with  = 0.25. | M1
A1
B1
A1
[4] | 2.1
1.1
1.1
2.2a | Correct method for probability
Correct formula
Simplified, any 
Correctly justify required result,
with  = 0.25 oe stated explicitly | 1 – e–0.25: M1A0
i.e. neither 0! nor e0 left in
Need to say “same” oe
PPMMTT
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
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
9 The continuous random variable $T$ has cumulative distribution function\\
$F ( t ) = \begin{cases} 0 & t < 0 , \\ 1 - \mathrm { e } ^ { - 0.25 t } & t \geqslant 0 . \end{cases}$
\begin{enumerate}[label=(\alph*)]
\item Find the cumulative distribution function of $2 T$.
\item Show that, for constant $k , \mathrm { E } \left( \mathrm { e } ^ { k t } \right) = \frac { 1 } { 1 - 4 k }$.

You should state with a reason the range of values of $k$ for which this result is valid.
\item $\quad T$ is the time before a certain event occurs.

Show that the probability that no event occurs between time $T = 0$ and time $T = \theta$ is the same as the probability that the value of a random variable with the distribution $\operatorname { Po } ( \lambda )$ is 0 , for a certain value of $\lambda$. You should state this value of $\lambda$ in terms of $\theta$.

\section*{END OF QUESTION PAPER}
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2019 Q9 [14]}}
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