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The probability density $\mathrm{f}(\mathrm{x})$ of a continuous random variable is given by $f(x)=$ $\mathrm{Ke}^{-|\mathrm{x}|},-\infty < \mathrm{x} < \infty$. Then the value of $\mathrm{K}$ is
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Since $\mathrm{f}(\mathrm{x})$ is the probability density function of random variable $\mathrm{X}$.
$\therefore \int_{-\infty}^{\infty} \mathrm{f}(\mathrm{x})=1$
Now we have
$\begin{array}{l}
\int_{-\infty}^{\infty} \mathrm{Ke}^{-|\mathrm{x}|} \mathrm{dx}=1 \Rightarrow 2 \int_{0}^{\infty} \mathrm{K} \cdot \mathrm{e}^{-|\mathrm{x}|} \mathrm{d} \mathrm{x}=1 \\
\Rightarrow 2 \int_{0}^{\infty} \mathrm{K} \cdot \mathrm{e}^{-\mathrm{x}} \mathrm{d} \mathrm{x}=1 \\
\Rightarrow-2 \mathrm{~K} \cdot\left[\mathrm{e}^{-\mathrm{x}}\right]_{0}^{\infty}=1 \Rightarrow 2 \mathrm{~K}=1 \\
\Rightarrow \mathrm{K}=\frac{1}{2}
\end{array}$
$\therefore \int_{-\infty}^{\infty} \mathrm{f}(\mathrm{x})=1$
Now we have
$\begin{array}{l}
\int_{-\infty}^{\infty} \mathrm{Ke}^{-|\mathrm{x}|} \mathrm{dx}=1 \Rightarrow 2 \int_{0}^{\infty} \mathrm{K} \cdot \mathrm{e}^{-|\mathrm{x}|} \mathrm{d} \mathrm{x}=1 \\
\Rightarrow 2 \int_{0}^{\infty} \mathrm{K} \cdot \mathrm{e}^{-\mathrm{x}} \mathrm{d} \mathrm{x}=1 \\
\Rightarrow-2 \mathrm{~K} \cdot\left[\mathrm{e}^{-\mathrm{x}}\right]_{0}^{\infty}=1 \Rightarrow 2 \mathrm{~K}=1 \\
\Rightarrow \mathrm{K}=\frac{1}{2}
\end{array}$
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