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Statistics, 2ADV S3 2024 MET2 14*

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)

Show Answers Only

\(k=\dfrac{1}{2}\)

Show Worked Solution

\(h(x)\ \text{is a PDF if}\ \ \displaystyle \int_{-3}^1 h(x)=1\)

\(\displaystyle \int_{-3}^1 h(x)\) \(=\displaystyle \int_{-3}^0 \dfrac{x}{6}+k\,dx +\int_0^1 -\dfrac{x}{2}+k\,dx\)
\(1\) \(=\left[\dfrac{x^2}{12}+kx\right] _{-3}^0 +\left[-\dfrac{x^2}{4}+kx\right] _0^1\) 
\(1\) \(=0-\left(\dfrac{9}{12}-3k\right)+\left(-\dfrac{1}{4}+k\right)-0\)
\(1\) \(=4k-1\)
\(k\) \(=\dfrac{1}{2}\)