OCR S3 2006 January — Question 5 12 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2006
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeMulti-part piecewise CDF
DifficultyStandard +0.8 This S3 question requires understanding of CDF properties (continuity, F(4)=1), finding percentiles, and transforming CDFs - a non-trivial transformation Y=√(X-1). While systematic, it demands careful handling of piecewise functions and the relationship between transformed variables, placing it moderately above average difficulty.
Spec5.03a Continuous random variables: pdf and cdf5.03f Relate pdf-cdf: medians and percentiles

5 The continuous random variable \(X\) has cumulative distribution function given by $$F ( x ) = \begin{cases} 0 & x < 1 , \\ \frac { 1 } { 8 } ( x - 1 ) ^ { 2 } & 1 \leqslant x < 3 , \\ a ( x - 2 ) & 3 \leqslant x < 4 , \\ 1 & x \geqslant 4 , \end{cases}$$ where \(a\) is a positive constant.
  1. Find the value of \(a\).
  2. Verify that \(C _ { X } ( 8 )\), the 8th percentile of \(X\), is 1.8 .
  3. Find the cumulative distribution function of \(Y\), where \(Y = \sqrt { X - 1 }\).
  4. Find \(C _ { Y } ( 8 )\) and verify that \(C _ { Y } ( 8 ) = \sqrt { C _ { X } ( 8 ) - 1 }\).

AnswerMarks Guidance
(i) EITHER: Use \(\frac{1}{3}(3-1)^2=a(3-2)\)M1 Continuity of \(F\)
OR: \(a(4-2)=1\)
\(a=\frac{1}{2}\)A1 2
(ii) \(F(1.8)=\frac{1}{3}(0.8)^2\)M1 Appropriate use of \(F\)
\(=0.08\)
\(C_X(8)=1.8\)A1 2
(iii) \(G(y)=P(Y\leq y)=P((X-1)^{1/2}\leq y)\)M1
\(=P((X-1)\leq y^2)\)
\(=P(X\leq y^2+1)\)A1
\(=F(y^2+1)\)
\[G(y)=\begin{cases}
\frac{1}{8}y^4 & (0 \leq y \leq \sqrt{2}), \\
\frac{1}{2}(y^2-1) & (\sqrt{2} < y \leq \sqrt{3}).
AnswerMarks Guidance
\end{cases}\]A1
Ignore others, A1 for both ranges of \(y\)B1 5
(iv) Use \(G(y)\) to find \(C_Y(8)\)M1
Obtain \(\sqrt{(0,8)}\)A1
Correct verificationB1 3
(i) EITHER: Use $\frac{1}{3}(3-1)^2=a(3-2)$ | M1 | Continuity of $F$
OR: $a(4-2)=1$ | |
$a=\frac{1}{2}$ | A1 | 2

(ii) $F(1.8)=\frac{1}{3}(0.8)^2$ | M1 | Appropriate use of $F$
$=0.08$ | |
$C_X(8)=1.8$ | A1 | 2

(iii) $G(y)=P(Y\leq y)=P((X-1)^{1/2}\leq y)$ | M1 |
$=P((X-1)\leq y^2)$ | |
$=P(X\leq y^2+1)$ | A1 |
$=F(y^2+1)$ | |
$$G(y)=\begin{cases}
\frac{1}{8}y^4 & (0 \leq y \leq \sqrt{2}), \\
\frac{1}{2}(y^2-1) & (\sqrt{2} < y \leq \sqrt{3}).
\end{cases}$$ | A1 |
Ignore others, A1 for both ranges of $y$ | B1 | 5

(iv) Use $G(y)$ to find $C_Y(8)$ | M1 |
Obtain $\sqrt{(0,8)}$ | A1 |
Correct verification | B1 | 3

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5 The continuous random variable $X$ has cumulative distribution function given by

$$F ( x ) = \begin{cases} 0 & x < 1 , \\ \frac { 1 } { 8 } ( x - 1 ) ^ { 2 } & 1 \leqslant x < 3 , \\ a ( x - 2 ) & 3 \leqslant x < 4 , \\ 1 & x \geqslant 4 , \end{cases}$$

where $a$ is a positive constant.\\
(i) Find the value of $a$.\\
(ii) Verify that $C _ { X } ( 8 )$, the 8th percentile of $X$, is 1.8 .\\
(iii) Find the cumulative distribution function of $Y$, where $Y = \sqrt { X - 1 }$.\\
(iv) Find $C _ { Y } ( 8 )$ and verify that $C _ { Y } ( 8 ) = \sqrt { C _ { X } ( 8 ) - 1 }$.

\hfill \mbox{\textit{OCR S3 2006 Q5 [12]}}