The probability density function on the interval [a, 1] is given by $1 / x^{2}$ and outside this interval the value of the function is zero. The value of a is

0.5

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Solution

The correct option is A 0.5
p.d.f. in $f (x) = (\begin{matrix} \frac{1}{x}, & a \leq n \leq 1 0, & othewise \end{matrix}$

for p.d.f we have

$\int_{- \infty}^{\infty} f (x) d x = 1$

$\Rightarrow \int_{a}^{1} f (x) d x = 1$

$\Rightarrow \int_{a}^{1} (\frac{1}{x^{2}}) d x = 1$

$\Rightarrow {[- \frac{1}{x}]}_{a}^{1} = 1 \Rightarrow a = \frac{1}{2}$

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