CAIE S2 2016 March — Question 7 11 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2016
SessionMarch
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeSymmetry property of PDF
DifficultyStandard +0.3 This question tests standard PDF properties including symmetry recognition for E(X), variance calculation via integration, and using symmetry for probability calculations. Part (b) involves routine median and normalization conditions. While requiring multiple techniques, all are textbook applications with no novel problem-solving—slightly easier than average due to the symmetry shortcut provided by the diagram.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03d E(g(X)): general expectation formula5.03f Relate pdf-cdf: medians and percentiles

7
  1. \includegraphics[max width=\textwidth, alt={}, center]{3f1a0c67-03a4-4b4f-99c0-4336ba7d56b0-3_255_643_264_790} The diagram shows the graph of the probability density function, f , of a random variable \(X\), where $$f ( x ) = \begin{cases} \frac { 2 } { 9 } \left( 3 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
    1. State the value of \(\mathrm { E } ( X )\) and find \(\operatorname { Var } ( X )\).
    2. State the value of \(\mathrm { P } ( 1.5 \leqslant X \leqslant 4 )\).
    3. Given that \(\mathrm { P } ( 1 \leqslant X \leqslant 2 ) = \frac { 13 } { 27 }\), find \(\mathrm { P } ( X > 2 )\).
  2. A random variable, \(W\), has probability density function given by $$\mathrm { g } ( w ) = \begin{cases} a w & 0 \leqslant w \leqslant b \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) and \(b\) are constants. Given that the median of \(W\) is 2 , find \(a\) and \(b\).

Question 7:
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(E(X) = 1.5\)B1
\(\frac{2}{9}\int_0^3 (3x^3 - x^4)\,dx\)M1 Attempt \(\int x^2 f(x)\), ignore limits
\(= \frac{2}{9}\left[\frac{3x^4}{4} - \frac{x^5}{5}\right]_0^3\)
\(= \frac{2}{9}\left[\frac{243}{4} - \frac{243}{5}\right]\) \((= 2.7)\)M1 Sub correct limits into correct integral
\(\text{Var}(X) = 2.7 - 1.5^2 = 0.45\) oeA1\(\checkmark\) [4] Ft their \(E(X)\), but no ft for negative Var
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(0.5\)B1 [1]
Part (a)(iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\left(1 - \frac{13}{27}\right) \div 2\)M1 or \(\frac{2}{9}\int_2^3 (3x - x^2)\,dx\) oe
\(= \frac{7}{27}\) or \(0.259\)A1 [2] As final answer
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{2} \times 2 \times 2a = \frac{1}{2}\) or \(\int_0^2 ax\,dx = \frac{1}{2}\)M1 Attempt correct equation in \(a\)
\(a = \frac{1}{4}\)A1
\(\frac{1}{2} \times b \times \frac{1}{4}b = 1\) or \(\int_0^b \frac{1}{4}x\,dx = 1\) or \(b = 2\times\sqrt{2}\)M1 or \(\frac{1}{2} \times b \times ab = 1\) or \(\int_0^b ax\,dx = 1\); attempt correct equation in \(a\) and \(b\)
\(b = 2\sqrt{2}\)A1\(\checkmark\) [4] Allow \(b = \sqrt{8}\) or 2.83 (3 sf); ft incorrect \(a\), both Ms needed
Total for paper: 50
## Question 7:

### Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = 1.5$ | **B1** | |
| $\frac{2}{9}\int_0^3 (3x^3 - x^4)\,dx$ | **M1** | Attempt $\int x^2 f(x)$, ignore limits |
| $= \frac{2}{9}\left[\frac{3x^4}{4} - \frac{x^5}{5}\right]_0^3$ | | |
| $= \frac{2}{9}\left[\frac{243}{4} - \frac{243}{5}\right]$ $(= 2.7)$ | **M1** | Sub correct limits into correct integral |
| $\text{Var}(X) = 2.7 - 1.5^2 = 0.45$ oe | **A1$\checkmark$** [4] | Ft their $E(X)$, but no ft for negative Var |

### Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5$ | **B1** [1] | |

### Part (a)(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(1 - \frac{13}{27}\right) \div 2$ | **M1** | or $\frac{2}{9}\int_2^3 (3x - x^2)\,dx$ oe |
| $= \frac{7}{27}$ or $0.259$ | **A1** [2] | As final answer |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 2 \times 2a = \frac{1}{2}$ or $\int_0^2 ax\,dx = \frac{1}{2}$ | **M1** | Attempt correct equation in $a$ |
| $a = \frac{1}{4}$ | **A1** | |
| $\frac{1}{2} \times b \times \frac{1}{4}b = 1$ or $\int_0^b \frac{1}{4}x\,dx = 1$ or $b = 2\times\sqrt{2}$ | **M1** | or $\frac{1}{2} \times b \times ab = 1$ or $\int_0^b ax\,dx = 1$; attempt correct equation in $a$ and $b$ |
| $b = 2\sqrt{2}$ | **A1$\checkmark$** [4] | Allow $b = \sqrt{8}$ or 2.83 (3 sf); ft incorrect $a$, both Ms needed |

---

**Total for paper: 50**
7
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{3f1a0c67-03a4-4b4f-99c0-4336ba7d56b0-3_255_643_264_790}

The diagram shows the graph of the probability density function, f , of a random variable $X$, where

$$f ( x ) = \begin{cases} \frac { 2 } { 9 } \left( 3 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item State the value of $\mathrm { E } ( X )$ and find $\operatorname { Var } ( X )$.
\item State the value of $\mathrm { P } ( 1.5 \leqslant X \leqslant 4 )$.
\item Given that $\mathrm { P } ( 1 \leqslant X \leqslant 2 ) = \frac { 13 } { 27 }$, find $\mathrm { P } ( X > 2 )$.
\end{enumerate}\item A random variable, $W$, has probability density function given by

$$\mathrm { g } ( w ) = \begin{cases} a w & 0 \leqslant w \leqslant b \\ 0 & \text { otherwise } \end{cases}$$

where $a$ and $b$ are constants. Given that the median of $W$ is 2 , find $a$ and $b$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2016 Q7 [11]}}