A probability density function on the interval \([a,1]\) is given by \(1/x^2\)…
2016
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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Correct answer: 0.5
Solution: Find a so that the probability density integrates to 1 over the interval [a, 1].
Compute the integral of the density over [a, 1].
∫_a^1 1/x^2 dx = [-1/x]_a^1 = -1 + 1/a = 1/a - 1.
Set this equal to 1 (total probability) and solve for a: 1/a - 1 = 1 ⇒ 1/a = 2 ⇒ a = 1/2.
Therefore the value of a is 1/2.