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A 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 _________.
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We know that the sum of all the probabilities is $1$

Therefore, on integrating $\frac{1}{x^2}$ with limits a to $1$, the result should be $1$.

Hence,  $\int_{a}^{1} \frac{1}{x^2} dx = 1$

            $\left[-\frac{1}{x}\right]_a^1 = 1$

            $-1 + \frac{1}{a} = 1$

Hence, a = 0.5

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The property of probability density function is area under curve = 1
or

where (a, b) is internal and f(x) is probability density function.
Given,
f(x) = 1/x2 , a≤x≤1
The area under curve,

- 1 + 1/a = 1
1/a = 2
 

Value of a=0.5

Answer:

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