The p.d.f. of a random variable \(\mathrm{X}\) is given by \(f(x)=\frac{k}{\sqrt{x}}\) if \(0 \leq x \leq 4\) \(=0\) otherwise, then \(\mathrm{P}(1 < X < 4)=\)

Given \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{K}}{\sqrt{\mathrm{x}}}, \) if \(0 \leq \mathrm{x} \leq 4\)
\(=0, \) otherwise
\(\therefore \int_{0}^{4} \frac{\mathrm{K}}{\sqrt{\mathrm{x}}} \mathrm{dx}=1 \Rightarrow \mathrm{K}\left[\frac{\mathrm{x}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{0}^{4}=1 \Rightarrow 2 \mathrm{~K}\) \((\sqrt{4}-0)=1\)
\(\therefore 4 \mathrm{~K}=1 \Rightarrow \mathrm{K}=\frac{1}{4}\)
\(\therefore \mathrm{P}(1 < \mathrm{x} < 4)=\int_{1}^{4}\left[\frac{\left(\frac{1}{4}\right)}{\sqrt{\mathrm{x}}}\right) \mathrm{dx}\)
\(=\frac{1}{4} \int_{1}^{4} \mathrm{x}^{-\frac{1}{2}} \mathrm{dx}=\frac{1}{4}\left[\frac{\mathrm{x}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{1}^{4}=\frac{2}{4}\left[4^{\frac{1}{2}}-1\right]=\frac{1}{2}\) \((2-1)=\frac{1}{2}\)