OCR Further Statistics Specimen — Question 9 9 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
SessionSpecimen
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeCDF of transformed variable
DifficultyChallenging +1.2 This is a Further Maths statistics question requiring transformation of CDFs using the inverse function method and an improper integral to show non-existence of expectation. While conceptually more advanced than A-level, the execution is relatively straightforward: finding P(Y≤y) = P(X≥1/√y), substituting into the given CDF, then showing E(Y) diverges by integrating x^(-3). The steps are mechanical once the method is recognized, making it moderately above average difficulty.
Spec5.03a Continuous random variables: pdf and cdf5.03g Cdf of transformed variables
9 The continuous random variable \(X\) has cumulative distribution function given by $$\mathrm { F } ( x ) = \left\{ \begin{array} { c c } 0 & x < 0 \\ \frac { 1 } { 16 } x ^ { 2 } & 0 \leq x \leq 4 \\ 1 & x > 4 \end{array} \right.$$
  1. The random variable \(Y\) is defined by \(Y = \frac { 1 } { X ^ { 2 } }\). Find the cumulative distribution function of \(Y\).
  2. Show that \(\mathrm { E } ( Y )\) is not defined. \section*{END OF QUESTION PAPER}
Question 9:
AnswerMarks Guidance
9(i) (cid:167) 1 (cid:183)
P(cid:11)Y (cid:100) y(cid:12)(cid:32)P (cid:168) (cid:100) y (cid:184)
(cid:169) X2 (cid:185)
(cid:167) 1 (cid:183)
(cid:32)P(cid:168)X (cid:116) (cid:184)
(cid:168) (cid:184)
y
(cid:169) (cid:185)
(cid:167) 1 (cid:183)
(cid:32)1(cid:16)F(cid:168) (cid:184)
(cid:168) (cid:184)
y
(cid:169) (cid:185)
(cid:173) 1 1
(cid:176)1(cid:16) y(cid:33) ,
(cid:32)(cid:174) 16y 16
(cid:176)
AnswerMarks
(cid:175) 0 otherwise.M1
E1
M1
E1
B1
AnswerMarks
[5]1.1a
2.1
2.1
3.1a
AnswerMarks
1.1Attempt to write F in terms of X
y
Make X the subject
1(cid:16)F(innverse function)
e
1
1(cid:16) correct, www
16y
m
AnswerMarks
0 and ranges correct (independent)Withhold if extra range(s)
given
AnswerMarks Guidance
9(ii) 1
PDF of y is
16y2
(cid:102)
(cid:180) y
(cid:181) dy
(cid:182)1 16y2
16
(cid:102)
(cid:32)(cid:170)1 lny(cid:186)
(cid:172)16 (cid:188)1
16
AnswerMarks
and lny is undefined as y(cid:111)(cid:102)M1
M1
A1
E1
AnswerMarks
[4]3.1a
1.1
2.1
AnswerMarks
3.2aDifferentiate CDF to find PDF of Y
Multiply by y and integrate, using their
limits
Integrantion must be shown explicitly
e
AnswerMarks
Correctly justify given statementA0A0 For “calculator gives
math error” or similar
AnswerMarks
ORPDF of x is 1x
8
(cid:167) 1 (cid:183)
E(cid:11)Y(cid:12)(cid:32)E
(cid:168) (cid:184)
(cid:169) X2 (cid:185)
(cid:180) 4 1
(cid:32)(cid:181) 1xdx
(cid:182) x2 8
0
(cid:32)(cid:170)1lnx(cid:186) 4 p
(cid:172)8 (cid:188)
0
AnswerMarks
and lnx is undefined as x(cid:111)0M1
c
M1
e
A1
E1
AnswerMarks Guidance
[4]i m
Differentiate CDF to find PDF of X
1
Integrate (cid:117)PDF, limits 0, 4
x2
Integration must be shown explicitly
AnswerMarks
Correctly justify given statementA0A0 For “calculator gives
math error” or similar
AnswerMarks Guidance
QuestionAO1 AO2
AnswerMarks Guidance
9(i)2 2
AnswerMarks Guidance
9(ii)1 1
Totals38 18
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Question 9:
9 | (i) | (cid:167) 1 (cid:183)
P(cid:11)Y (cid:100) y(cid:12)(cid:32)P (cid:168) (cid:100) y (cid:184)
(cid:169) X2 (cid:185)
(cid:167) 1 (cid:183)
(cid:32)P(cid:168)X (cid:116) (cid:184)
(cid:168) (cid:184)
y
(cid:169) (cid:185)
(cid:167) 1 (cid:183)
(cid:32)1(cid:16)F(cid:168) (cid:184)
(cid:168) (cid:184)
y
(cid:169) (cid:185)
(cid:173) 1 1
(cid:176)1(cid:16) y(cid:33) ,
(cid:32)(cid:174) 16y 16
(cid:176)
(cid:175) 0 otherwise. | M1
E1
M1
E1
B1
[5] | 1.1a
2.1
2.1
3.1a
1.1 | Attempt to write F in terms of X
y
Make X the subject
1(cid:16)F(innverse function)
e
1
1(cid:16) correct, www
16y
m
0 and ranges correct (independent) | Withhold if extra range(s)
given
9 | (ii) | 1
PDF of y is
16y2
(cid:102)
(cid:180) y
(cid:181) dy
(cid:182)1 16y2
16
(cid:102)
(cid:32)(cid:170)1 lny(cid:186)
(cid:172)16 (cid:188)1
16
and lny is undefined as y(cid:111)(cid:102) | M1
M1
A1
E1
[4] | 3.1a
1.1
2.1
3.2a | Differentiate CDF to find PDF of Y
Multiply by y and integrate, using their
limits
Integrantion must be shown explicitly
e
Correctly justify given statement | A0A0 For “calculator gives
math error” or similar
OR | PDF of x is 1x
8
(cid:167) 1 (cid:183)
E(cid:11)Y(cid:12)(cid:32)E
(cid:168) (cid:184)
(cid:169) X2 (cid:185)
(cid:180) 4 1
(cid:32)(cid:181) 1xdx
(cid:182) x2 8
0
(cid:32)(cid:170)1lnx(cid:186) 4 p
(cid:172)8 (cid:188)
0
and lnx is undefined as x(cid:111)0 | M1
c
M1
e
A1
E1
[4] | i | m
Differentiate CDF to find PDF of X
1
Integrate (cid:117)PDF, limits 0, 4
x2
Integration must be shown explicitly
Correctly justify given statement | A0A0 For “calculator gives
math error” or similar
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Total
--- 9(i) ---
9(i) | 2 | 2 | 10 | c | 5
--- 9(ii) ---
9(ii) | 1 | 1 | 20 | e | 4
Totals | 38 | 18 | 8 | 11 | 75
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9 The continuous random variable $X$ has cumulative distribution function given by

$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c } 
0 & x < 0 \\
\frac { 1 } { 16 } x ^ { 2 } & 0 \leq x \leq 4 \\
1 & x > 4
\end{array} \right.$$

(i) The random variable $Y$ is defined by $Y = \frac { 1 } { X ^ { 2 } }$. Find the cumulative distribution function of $Y$.\\
(ii) Show that $\mathrm { E } ( Y )$ is not defined.

\section*{END OF QUESTION PAPER}

\hfill \mbox{\textit{OCR Further Statistics  Q9 [9]}}
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