CAIE S2 2019 June — Question 4 9 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2019
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeDirect variance calculation from pdf
DifficultyStandard +0.3 This is a straightforward S2 question requiring standard pdf techniques: using the total probability property to find the constant, calculating E(X) and Var(X) using integration, and solving a cumulative probability equation. All steps are routine applications of formulas with no conceptual challenges or novel problem-solving required. Slightly easier than average due to the simple uniform distribution in part (a) and straightforward integration in part (b).
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration

4

  1. The diagram shows the graph of the probability density function, f , of a random variable \(X\), where \(a\) is a constant greater than 0.5 . The graph between \(x = 0\) and \(x = a\) is a straight line parallel to the \(x\)-axis.
    1. Find \(\mathrm { P } ( X < 0.5 )\) in terms of \(a\).
    2. Find \(\mathrm { E } ( X )\) in terms of \(a\).
    3. Show that \(\operatorname { Var } ( X ) = \frac { 1 } { 12 } a ^ { 2 }\).
  2. A random variable \(T\) has probability density function given by $$\operatorname { g } ( t ) = \begin{cases} \frac { 3 } { 2 ( t - 1 ) ^ { 2 } } & 2 \leqslant t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ Find the value of \(b\) such that \(\mathrm { P } ( T \leqslant b ) = \frac { 3 } { 4 }\).

Question 4(a)(i):
AnswerMarks Guidance
AnswerMark Guidance
\(0.5 \times \frac{1}{a} = \left(\frac{0.5}{a}\right)\)M1 Or attempt to integrate \(f(x)\) (\(=1/a\)) between 0 and 0.5
\(= \frac{1}{2a}\) oeA1 Accept \(0.5/a\) for A1
Total: 2
Question 4(a)(ii):
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{a}{2}\)B1
Total: 1
Question 4(a)(iii):
AnswerMarks Guidance
AnswerMark Guidance
\(\int_0^a \frac{x^2}{a}\,dx - \left('\frac{a}{2}'\right)^2\)M1 Integrate \(x^2 f(x)\) from 0 to \(a\) and sub their mean\(^2\)
\(\text{Var}(X) = \frac{a^2}{3} - \frac{a^2}{4}\)A1 Must see this line oe
\(\left(\text{Var}(X) = \frac{a^2}{12}\right)\) AG
Total: 2
Question 4(b):
AnswerMarks Guidance
AnswerMark Guidance
\(\int_2^b \frac{3}{2(t-1)^2}\,dt\)M1 Attempt to integrate \(g(t)\), ignore limits
\(\left[-\frac{3}{2(t-1)}\right]_2^b\)A1 Correct integral
\(-\frac{3}{2}\left(\frac{1}{(b-1)}-1\right) = \frac{3}{4}\); \(\left(1 - \frac{1}{(b-1)} = \frac{1}{2}\right)\)M1 Attempt subst correct limits in their integral and \(= \frac{3}{4}\)
\(b = 3\)A1
Total: 4
## Question 4(a)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $0.5 \times \frac{1}{a} = \left(\frac{0.5}{a}\right)$ | M1 | Or attempt to integrate $f(x)$ ($=1/a$) between 0 and 0.5 |
| $= \frac{1}{2a}$ oe | A1 | Accept $0.5/a$ for A1 |
| **Total: 2** | | |

---

## Question 4(a)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{a}{2}$ | B1 | |
| **Total: 1** | | |

---

## Question 4(a)(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^a \frac{x^2}{a}\,dx - \left('\frac{a}{2}'\right)^2$ | M1 | Integrate $x^2 f(x)$ from 0 to $a$ and sub their mean$^2$ |
| $\text{Var}(X) = \frac{a^2}{3} - \frac{a^2}{4}$ | A1 | Must see this line oe |
| $\left(\text{Var}(X) = \frac{a^2}{12}\right)$ **AG** | | |
| **Total: 2** | | |

---

## Question 4(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_2^b \frac{3}{2(t-1)^2}\,dt$ | M1 | Attempt to integrate $g(t)$, ignore limits |
| $\left[-\frac{3}{2(t-1)}\right]_2^b$ | A1 | Correct integral |
| $-\frac{3}{2}\left(\frac{1}{(b-1)}-1\right) = \frac{3}{4}$; $\left(1 - \frac{1}{(b-1)} = \frac{1}{2}\right)$ | M1 | Attempt subst correct limits in their integral and $= \frac{3}{4}$ |
| $b = 3$ | A1 | |
| **Total: 4** | | |

---
4
\begin{enumerate}[label=(\alph*)]
\item \\
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The diagram shows the graph of the probability density function, f , of a random variable $X$, where $a$ is a constant greater than 0.5 . The graph between $x = 0$ and $x = a$ is a straight line parallel to the $x$-axis.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { P } ( X < 0.5 )$ in terms of $a$.
\item Find $\mathrm { E } ( X )$ in terms of $a$.
\item Show that $\operatorname { Var } ( X ) = \frac { 1 } { 12 } a ^ { 2 }$.
\end{enumerate}\item A random variable $T$ has probability density function given by

$$\operatorname { g } ( t ) = \begin{cases} \frac { 3 } { 2 ( t - 1 ) ^ { 2 } } & 2 \leqslant t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$

Find the value of $b$ such that $\mathrm { P } ( T \leqslant b ) = \frac { 3 } { 4 }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2019 Q4 [9]}}