OCR Further Statistics 2023 June — Question 6 7 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeCDF of transformed variable
DifficultyChallenging +1.8 This requires finding the CDF of a non-linear transformation (sine) of a uniform variable, involving careful consideration of the inverse function's multi-valued nature over [-π, π] and integration. It demands strong understanding of transformations and geometric insight about the sine function's behavior, going well beyond routine CDF calculations.
Spec5.03g Cdf of transformed variables

6 The continuous random variable \(X\) has a uniform distribution on the interval \([ - \pi , \pi ]\).
The random variable \(Y\) is defined by \(Y = \sin X\).
Determine the cumulative distribution function of \(Y\).

Question 6:
AnswerMarks
6π+x
CDF of X is (F(x) =)
P(Y < y) = P(sin X < y) (= P(X < sin–1 y))
s i n − 1 y
+
AnswerMarks
2 πM1
A1
M1
AnswerMarks
M12.1
1.1
2.1
AnswerMarks
2.1Attempt to find CDF of X, e.g. x/2
Correct
Allow for F(sin–1 y), and allow inequality errors
Method for dealing with ranges
AnswerMarks Guidance
ORP(Y < y) = P(sin X < y) M1
= F(sin–1 y) where F is CDF of XM1
Method for dealing with rangesM1
+ ½A1 OE
π+2sin−1y
Range of Y is –1  y  1
 0 y−1,
π+2sin−1y
F(y)= −1 y1,
 2π
 1 y1.
AnswerMarks
A1
B1
A1ft
AnswerMarks
[7]2.2a
1.1
AnswerMarks
1.2Any equivalent form, WWW
Stated. NB –  sin–1 y   is insufficient, same letter as in function
Correct including 0 and 1, with ranges, but allow = in wrong places, FT
on their π + 2 s in − 1 y . Must be in terms of y (if F(y), or x if F(x), etc)
2 π
(not – < sin–1 y <  etc), depends on second M1
Question 6:
6 | π+x
CDF of X is (F(x) =)
2π
P(Y < y) = P(sin X < y) (= P(X < sin–1 y))
s i n − 1 y
+
2 π | M1
A1
M1
M1 | 2.1
1.1
2.1
2.1 | Attempt to find CDF of X, e.g. x/2
Correct
Allow for F(sin–1 y), and allow inequality errors
Method for dealing with ranges
OR | P(Y < y) = P(sin X < y) | M1
= F(sin–1 y) where F is CDF of X | M1
Method for dealing with ranges | M1
+ ½ | A1 | OE
π+2sin−1y
2π
Range of Y is –1  y  1
 0 y−1,

π+2sin−1y
F(y)= −1 y1,
 2π
 1 y1.
 | A1
B1
A1ft
[7] | 2.2a
1.1
1.2 | Any equivalent form, WWW
Stated. NB –  sin–1 y   is insufficient, same letter as in function
Correct including 0 and 1, with ranges, but allow = in wrong places, FT
on their π + 2 s in − 1 y . Must be in terms of y (if F(y), or x if F(x), etc)
2 π
(not – < sin–1 y <  etc), depends on second M1
6 The continuous random variable $X$ has a uniform distribution on the interval $[ - \pi , \pi ]$.\\
The random variable $Y$ is defined by $Y = \sin X$.\\
Determine the cumulative distribution function of $Y$.

\hfill \mbox{\textit{OCR Further Statistics 2023 Q6 [7]}}