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
DifficultyChallenging +1.2 This question tests understanding of symmetry in probability distributions and requires integration of a polynomial PDF. Part (a) involves algebraic manipulation with symmetry properties, part (b) applies symmetry directly, and part (c) requires setting up and solving an integral equation. While it involves multiple steps and conceptual understanding of symmetry, the techniques are standard S2 material with straightforward integration and algebra—moderately above average difficulty but not requiring novel insight.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles
6 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 \(\mathrm { P } ( a < X < b ) = p\) and \(\mathrm { P } ( b < X < 3 ) = \frac { 13 } { 10 } p\), where \(p\) is a positive constant.
  1. Show that \(p \leqslant \frac { 5 } { 23 }\).
  2. Find \(\mathrm { P } ( b < X < 6 - a )\) in terms of \(p\).
    It is now given that the probability density function of \(X\) is f , where $$f ( x ) = \begin{cases} \frac { 1 } { 36 } \left( 6 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$
  3. Given that \(b = 2\) and \(p = \frac { 5 } { 27 }\), find the value of \(a\).
Question 6(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(p + \frac{13}{10}p \leq \frac{1}{2} \Rightarrow p \leq \frac{5}{23}\) AGB1 Allow '=' in working but need an inequality in the answer. Allow \(0 < p \leq \frac{5}{23}\)
Question 6(b):
AnswerMarks Guidance
AnswerMarks Guidance
e.g. \(0.5 - 2.3p\) (0 to \(a\)), \(p + 1.3p\) (\(a\) to 3), \(2 \times 1.3p\) (\(2 \times (b\) to 3)), \(2.3p + 1.3p\) (\(a\) to \(3 + b\) to 3), \(2p + 2.6p\) (\(2 \times (a\) to 3)), \(0.5 - 1.3p\) (0 to \(b\)), \(0.5 + 1.3p\) (\(b\) to 6)M1 Any correct expression for the probability of a relevant region
\(\frac{18}{5}p\) or \(3.6p\)A1
Question 6(c):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{36}\int_{a}^{2}(6x - x^2)\,dx = \frac{5}{27}\)M1 Attempt to integrate with correct limits and equate to \(\frac{5}{27}\), oe. Integrate from 2 to \(6-a\) and equate to \(\frac{18}{5}p = \frac{2}{3}\). Integrate from \(a\) to 3 and equate to \(\frac{23}{54}\). Integrate from 0 to \(a\) and equate to \(\frac{2}{27}\)
\(\Rightarrow \frac{1}{36}\left[3x^2 - \frac{x^3}{3}\right]_{a}^{2} = \frac{5}{27} \Rightarrow \frac{1}{36}\left(12 - \frac{8}{3} - 3a^2 + \frac{a^3}{3}\right) = \frac{5}{27}\)M1 For integrating and substitution of limits to form cubic in \(a\)
\(a^3 - 9a^2 + 8 = 0\)A1 Any correct three term cubic equation in \(a\)
\((a-1)(a^2 - 8a - 8) = 0\)M1 Attempt to factorise their cubic equation
\(a = \frac{8 \pm \sqrt{96}}{2} = 4 \pm \sqrt{24}\) or \(-0.899\) or \(8.90\), [not between 0 and 6]
\(a = 1\) only [other two values rejected]A1 SC B1 for \(a = 1\) only, if no method seen for solving the cubic 
## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $p + \frac{13}{10}p \leq \frac{1}{2} \Rightarrow p \leq \frac{5}{23}$ AG | B1 | Allow '=' in working but need an inequality in the answer. Allow $0 < p \leq \frac{5}{23}$ |

## Question 6(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. $0.5 - 2.3p$ (0 to $a$), $p + 1.3p$ ($a$ to 3), $2 \times 1.3p$ ($2 \times (b$ to 3)), $2.3p + 1.3p$ ($a$ to $3 + b$ to 3), $2p + 2.6p$ ($2 \times (a$ to 3)), $0.5 - 1.3p$ (0 to $b$), $0.5 + 1.3p$ ($b$ to 6) | M1 | Any correct expression for the probability of a relevant region |
| $\frac{18}{5}p$ or $3.6p$ | A1 | |

## Question 6(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{36}\int_{a}^{2}(6x - x^2)\,dx = \frac{5}{27}$ | M1 | Attempt to integrate with correct limits and equate to $\frac{5}{27}$, oe. Integrate from 2 to $6-a$ and equate to $\frac{18}{5}p = \frac{2}{3}$. Integrate from $a$ to 3 and equate to $\frac{23}{54}$. Integrate from 0 to $a$ and equate to $\frac{2}{27}$ |
| $\Rightarrow \frac{1}{36}\left[3x^2 - \frac{x^3}{3}\right]_{a}^{2} = \frac{5}{27} \Rightarrow \frac{1}{36}\left(12 - \frac{8}{3} - 3a^2 + \frac{a^3}{3}\right) = \frac{5}{27}$ | M1 | For integrating and substitution of limits to form cubic in $a$ |
| $a^3 - 9a^2 + 8 = 0$ | A1 | Any correct three term cubic equation in $a$ |
| $(a-1)(a^2 - 8a - 8) = 0$ | M1 | Attempt to factorise their cubic equation |
| $a = \frac{8 \pm \sqrt{96}}{2} = 4 \pm \sqrt{24}$ or $-0.899$ or $8.90$, [not between 0 and 6] | | |
| $a = 1$ only [other two values rejected] | A1 | **SC B1** for $a = 1$ only, if no method seen for solving the cubic |
6 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 $\mathrm { P } ( a < X < b ) = p$ and $\mathrm { P } ( b < X < 3 ) = \frac { 13 } { 10 } p$, where $p$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $p \leqslant \frac { 5 } { 23 }$.
\item Find $\mathrm { P } ( b < X < 6 - a )$ in terms of $p$.\\

It is now given that the probability density function of $X$ is f , where

$$f ( x ) = \begin{cases} \frac { 1 } { 36 } \left( 6 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$
\item Given that $b = 2$ and $p = \frac { 5 } { 27 }$, find the value of $a$.
\end{enumerate}

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