Aussie Maths & Science Teachers: Save your time with SmarterEd
A function, \(h(x)\), is defined as
\(h(x)=\left\{
\begin{array} {c}
\rule{0pt}{2.5ex} \ \ \ \ \ \dfrac{x}{6}+k \rule[-1ex]{0pt}{0pt} & -3 \leq x<0 \\
\rule{0pt}{2.5ex} \ \ -\dfrac{x}{2}+k \rule[-1ex]{0pt}{0pt} & 0 \leq x \leq 1 \\
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & \text { elsewhere } \\
\end{array}\right.\)
and \(k\) is a constant.
Find the value of \(k\) such that \(h(x)\) is a probability density function. (3 marks)
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A function \(f(x)\) is defined as
\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
1-\dfrac{x}{h}, & \text { for}\ \ 0 \leq x \leq h, \\
0, & \text { for}\ \ x \gt h \end{array}\right.\)
where \(h\) is a constant.
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a. \(h=2\)
b.
c. \(\text{Median}\ =0.586\)
a. \(f(x)\ \text{is a PDF if:}\)
| \(\displaystyle \int_{0}^{h} 1-\dfrac{x}{h}\,dx\) | \(=1\) |
| \(\Big[x-\dfrac{x^2}{2h} \Big]_0^{h}\) | \(=1\) |
| \(h-\dfrac{h}{2}\) | \(=1\) |
| \(h\) | \(=2\) |
b. \( \displaystyle \int 1-\dfrac{x}{2}\,dx = x-\dfrac{x^2}{4} + c\)
\(\text{At}\ \ x=0, \ \ F(0)=0,\ \ c=0 \)
\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
x-\dfrac{x^2}{4}, & \text { for}\ \ 0 \leq x \leq 2, \\
1, & \text { for}\ \ x \gt 2 \end{array}\right.\)
c. \(\text{Let}\ \ m=\ \text{median of}\ f(x) \)
| \(\displaystyle \int_0^{m} 1-\dfrac{x}{2}\,dx\) | \(=0.5\) |
| \(\Big[ x-\dfrac{x^2}{4} \Big]_0^m\) | \(=0.5\) |
| \(m-\dfrac{m^2}{4}\) | \(=0.5\) |
| \(4m-m^2\) | \(=2\) |
| \(m^2-4m+2\) | \(=0\) |
| \((m-2)^2\) | \(=2\) |
| \(m\) | \(=2 \pm \sqrt{2}\) |
| \(\therefore \ \text{Median}\) | \(=2-\sqrt{2}\ \ (x \in [0,2]) \) |
| \(=0.586\ \text{(3 d.p.)}\) |
A continuous random variable `X` has a probability density function given by
`f(x) = {{:(Cx + D),(0):}\ \ \ \ {:(2 <= x <= 5),(text(elsewhere)):}:}`
where `C` and `D` are constants.
Find the exact values of `C` and `D`, given the median of `X` is 4. (4 marks)
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The time Jennifer spends on her homework each day varies, but she does some homework every day.
The continuous random variable \(T\), which models the time, \(t\), in minutes, that Jennifer spends each day on her homework, has a probability density function \(f\), where
\(f(t)= \begin{cases}
\dfrac{1}{625}(t-20) & 20 \leq t<45 \\
\ \\
\dfrac{1}{625}(70-t) & 45 \leq t \leq 70 \\
\ \\
0 & \text {elsewhere }
\end{cases}\)
| a. |  |
MARKER’S COMMENT: Many did not draw graph along \(t\)-axis between 0 and 20 and for \(t>70\).
b. \(\text{Mode \(=45\) (value of \(t\) at highest value of \(f(t))\)}\)
c. \(P(25 \leq T \leq 55)\)
\(\begin{aligned}
& =\int_{25}^{45} \dfrac{1}{625}(t-20) d t+\int_{45}^{55} \dfrac{1}{625}(70-t) d t \\
& =\dfrac{1}{625}\left[\dfrac{t^2}{2}-20 t\right]_{25}^{45}+\dfrac{1}{625}\left[70 t-\dfrac{t^2}{2}\right]_{45}^{55} \\
& =\dfrac{1}{625}[112.5-(-187.5)]+\dfrac{1}{625}(2337.5-2137.5) \\ & =\dfrac{300}{625}+\dfrac{200}{625} \\
& =\dfrac{4}{5}
\end{aligned}\)
The probability density function of a continuous random variable \(X\) is given by
\(f(x)=\begin{cases}
\dfrac{x}{12} & 1 \leq x \leq 5 \\
\ \\
0 & \text {otherwise }
\end{cases}\)
Find \(P(X < 3)\) (2 marks)
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The probability density function \(f(x)\) of a random variable \(X\) is given by
\(f(x)=\begin{cases}
\dfrac{x+1}{12} & 0 \leq x \leq 4 \\
\ \\
0 & \text{otherwise }
\end{cases}\)
Find the value of \(b\) such that \(P(X \leq b)=\dfrac{5}{8}\). (3 marks)
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If a continuous random variable \(X\) has probability density function
\(f(x)=
\begin{cases}
\dfrac{x}{2} & \text{if } \quad 0 \leq x \leq 2 \\
\ \\
0 & \text {otherwise }
\end{cases}\)
Find the exact value of \(p\) such that \(P(X>p) = 0.4\) (3 marks)
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