CAIE S2 2023 November — Question 6 8 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2023
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeSymmetry property of PDF
DifficultyStandard +0.8 This S2 question requires understanding of symmetry in probability distributions, manipulation of probability statements, and solving a cubic equation from integrating a quadratic pdf. Part (c) involves setting up and solving ∫f(x)dx = 5/81 which leads to non-trivial algebraic manipulation. The multi-step reasoning and integration of several concepts (symmetry, pdf integration, equation solving) makes this moderately challenging but still within standard A-level techniques.
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
A continuous random variable \(X\) takes values from 0 to 6 only and has a probability distribution that is symmetrical. Two values, \(a\) and \(b\), of \(X\) are such that P\((a < X < b) = p\) and P\((b < X < 3) = \frac{13}{10}p\), where \(p\) is a positive constant.
  1. Show that \(p \leq \frac{5}{23}\). [1]
  2. Find P\((b < X < 6 - a)\) in terms of \(p\). [2]
It is now given that the probability density function of \(X\) is \(f\), where $$f(x) = \begin{cases} \frac{1}{36}(6x - x^2) & 0 \leq x \leq 6, \\ 0 & \text{otherwise}. \end{cases}$$
  1. Given that \(b = 2\) and \(p = \frac{5}{81}\), find the value of \(a\). [5]
Question 6:
AnswerMarks
6(a)13 1 5
p + p ⩽  p ⩽ AG
AnswerMarks Guidance
10 2 23B1 Allow ‘=’ in working but need an inequality in the
answer.
5
Allow 0 < p ⩽ .
23
1
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)e.g. 0.5 − 2.3p, p + 1.3p, 2 × 1.3p, 2.3p + 1.3p,
0 to a a to 3 2 × (b to 3) a to 3 + b to 3
2p + 2.6p, 0.5 − 1.3p, 0.5 + 1.3p,
AnswerMarks Guidance
2 × (a to 3) 0 to b b to 6M1 Any correct expression for the probability of a relevant
region.
18
por 3.6p
AnswerMarks
5A1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(c)2
1 (6x−x2)dx = 5
36 27
AnswerMarks Guidance
aM1 Attempt to integrate with correct limits and equate to
5
,oe.
27
18 2
Integrate from 2 to 6 – a and equate p= .
5 3
23
Integrate from a to 3 and equate to .
54
2
Integrate from 0 to a and equate to .
27
AnswerMarks Guidance
    3 1 6   3x2 − x 3 3   a 2 = 2 5 7  3 1 6    12− 8 3 −3a2 + a 3 3    = 2 5 7  M1 For integrating and substitution of limits to form cubic
in a.
AnswerMarks Guidance
a3 – 9a2 + 8 = 0A1 Any correct three term cubic equation in a.
(a – 1)(a2 – 8a – 8) = 0M1 Attempt to factorise their cubic equation.
8 96
a = = 4 ± 24 or –0.899 or 8.90, [not between 0 and 6]
2
AnswerMarks Guidance
a = 1 only [other two values rejected]A1 SC B1 for a = 1 only, if no method seen for solving the
cubic.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | 13 1 5
p + p ⩽  p ⩽ AG
10 2 23 | B1 | Allow ‘=’ in working but need an inequality in the
answer.
5
Allow 0 < p ⩽ .
23
1
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | e.g. 0.5 − 2.3p, p + 1.3p, 2 × 1.3p, 2.3p + 1.3p,
0 to a a to 3 2 × (b to 3) a to 3 + b to 3
2p + 2.6p, 0.5 − 1.3p, 0.5 + 1.3p,
2 × (a to 3) 0 to b b to 6 | M1 | Any correct expression for the probability of a relevant
region.
18
por 3.6p
5 | A1
2
Question | Answer | Marks | Guidance
--- 6(c) ---
6(c) | 2
1 (6x−x2)dx = 5
36 27
a | M1 | Attempt to integrate with correct limits and equate to
5
,oe.
27
18 2
Integrate from 2 to 6 – a and equate p= .
5 3
23
Integrate from a to 3 and equate to .
54
2
Integrate from 0 to a and equate to .
27
    3 1 6   3x2 − x 3 3   a 2 = 2 5 7  3 1 6    12− 8 3 −3a2 + a 3 3    = 2 5 7   | M1 | For integrating and substitution of limits to form cubic
in a.
a3 – 9a2 + 8 = 0 | A1 | Any correct three term cubic equation in a.
(a – 1)(a2 – 8a – 8) = 0 | M1 | Attempt to factorise their cubic equation.
8 96
a = = 4 ± 24 or –0.899 or 8.90, [not between 0 and 6]
2
a = 1 only [other two values rejected] | A1 | SC B1 for a = 1 only, if no method seen for solving the
cubic.
5
Question | Answer | Marks | Guidance
A continuous random variable $X$ takes values from 0 to 6 only and has a probability distribution that is symmetrical.

Two values, $a$ and $b$, of $X$ are such that P$(a < X < b) = p$ and P$(b < X < 3) = \frac{13}{10}p$, where $p$ is a positive constant.

\begin{enumerate}[label=(\alph*)]
\item Show that $p \leq \frac{5}{23}$. [1]

\item Find P$(b < X < 6 - a)$ in terms of $p$. [2]
\end{enumerate}

It is now given that the probability density function of $X$ is $f$, where
$$f(x) = \begin{cases} 
\frac{1}{36}(6x - x^2) & 0 \leq x \leq 6, \\
0 & \text{otherwise}.
\end{cases}$$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Given that $b = 2$ and $p = \frac{5}{81}$, find the value of $a$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2023 Q6 [8]}}
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