CAIE S2 2021 June — Question 6 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeDirect variance calculation from pdf
DifficultyStandard +0.3 This is a standard continuous probability distribution question requiring routine integration to find k, then E(X) and Var(X). The symmetry of the quadratic function allows E(X) to be stated immediately (midpoint = 3), and the variance calculation follows standard formulas with straightforward polynomial integration. Slightly easier than average due to the symmetric property and mechanical nature of the calculations.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration

The probability density function, f, of a random variable \(X\) is given by $$\text{f}(x) = \begin{cases} k(6x - x^2) & 0 \leq x \leq 6, \\ 0 & \text{otherwise,} \end{cases}$$ where \(k\) is a constant. State the value of \(\text{E}(X)\) and show that \(\text{Var}(X) = \frac{9}{5}\). [6]

Question 6:
AnswerMarks
6Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PPMMTT
9709/61 Cambridge International A Level – Mark Scheme May/June 2021
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2021 Page 5 of 13
AnswerMarks Guidance
QuestionAnswer Marks
6E(X) = 3 B1
6
k(6x−x2)dx = 1
0
 x36
k3x2 −  [= 1]
3 0
AnswerMarks Guidance
 M1 Attempt integration of f(x) and =1.
Ignore limits at this stage.
 216
k  108−  = 1
 3 
3 1
k = or
AnswerMarks
108 36A1
6
3
' '(6x3 −x4)dx
108
0
3 3x4 x56
=  −  = 10.8
108 2 5 0
AnswerMarks Guidance
 *M1 Attempt integration of their k × x2f(x).
Ignore limits at this stage. Accept in terms of k.
AnswerMarks Guidance
'10.8' – '3'2DM1 Their 10.8 (from use of limits 0 and 6) minus their
(E(X))2.
Accept in terms of k: 388.8k–(108k)2
9
or 1.8
AnswerMarks Guidance
5A1 CWO. Must be convincingly obtained as AG.
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PPMMTT
9709/61 Cambridge International A Level – Mark Scheme May/June 2021
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2021 Page 5 of 13
Question | Answer | Marks | Guidance
6 | E(X) = 3 | B1 | N.B. E(X)=108k is B0 until correct k substituted in.
6
k(6x−x2)dx = 1
0
 x36
k3x2 −  [= 1]
3 0
  | M1 | Attempt integration of f(x) and =1.
Ignore limits at this stage.
 216
k  108−  = 1
 3 
3 1
k = or
108 36 | A1
6
3
' '(6x3 −x4)dx
108
0
3 3x4 x56
=  −  = 10.8
108 2 5 0
  | *M1 | Attempt integration of their k × x2f(x).
Ignore limits at this stage. Accept in terms of k.
'10.8' – '3'2 | DM1 | Their 10.8 (from use of limits 0 and 6) minus their
(E(X))2.
Accept in terms of k: 388.8k–(108k)2
9
or 1.8
5 | A1 | CWO. Must be convincingly obtained as AG.
6
Question | Answer | Marks | Guidance
The probability density function, f, of a random variable $X$ is given by
$$\text{f}(x) = \begin{cases} k(6x - x^2) & 0 \leq x \leq 6, \\ 0 & \text{otherwise,} \end{cases}$$
where $k$ is a constant.

State the value of $\text{E}(X)$ and show that $\text{Var}(X) = \frac{9}{5}$. [6]

\hfill \mbox{\textit{CAIE S2 2021 Q6 [6]}}