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$\quad$ The p.d.f. of a random variable $\mathrm{X}$ is given by $f(x)=\frac{k}{\sqrt{x}} \quad$ if $0 \leq x \leq 4$
$=0$
otherwise, then $\mathrm{P}(1 < X < 4)=$
Options:
$=0$
otherwise, then $\mathrm{P}(1 < X < 4)=$
Solution:
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Given $\mathrm{f}(\mathrm{x})=\frac{\mathrm{K}}{\sqrt{\mathrm{x}}}, \quad$ if $0 \leq \mathrm{x} \leq 4$
$\quad=0, \quad$ otherwise
$\therefore \quad \int_{0}^{4} \frac{\mathrm{K}}{\sqrt{\mathrm{x}}} \mathrm{dx}=1 \Rightarrow \mathrm{K}\left[\frac{\mathrm{x}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{0}^{4}=1 \Rightarrow 2 \mathrm{~K}(\sqrt{4}-0)=1$
$\therefore \quad 4 \mathrm{~K}=1 \Rightarrow \mathrm{K}=\frac{1}{4}$
$\therefore \mathrm{P}(1 < \mathrm{x} < 4)=\int_{1}^{4}\left[\frac{\left(\frac{1}{4}\right)}{\sqrt{\mathrm{x}}}\right) \mathrm{dx}$
$=\frac{1}{4} \int_{1}^{4} \mathrm{x}^{-\frac{1}{2}} \mathrm{dx}=\frac{1}{4}\left[\frac{\mathrm{x}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{1}^{4}=\frac{2}{4}\left[4^{\frac{1}{2}}-1\right]=\frac{1}{2}(2-1)=\frac{1}{2}$
$\quad=0, \quad$ otherwise
$\therefore \quad \int_{0}^{4} \frac{\mathrm{K}}{\sqrt{\mathrm{x}}} \mathrm{dx}=1 \Rightarrow \mathrm{K}\left[\frac{\mathrm{x}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{0}^{4}=1 \Rightarrow 2 \mathrm{~K}(\sqrt{4}-0)=1$
$\therefore \quad 4 \mathrm{~K}=1 \Rightarrow \mathrm{K}=\frac{1}{4}$
$\therefore \mathrm{P}(1 < \mathrm{x} < 4)=\int_{1}^{4}\left[\frac{\left(\frac{1}{4}\right)}{\sqrt{\mathrm{x}}}\right) \mathrm{dx}$
$=\frac{1}{4} \int_{1}^{4} \mathrm{x}^{-\frac{1}{2}} \mathrm{dx}=\frac{1}{4}\left[\frac{\mathrm{x}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{1}^{4}=\frac{2}{4}\left[4^{\frac{1}{2}}-1\right]=\frac{1}{2}(2-1)=\frac{1}{2}$
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