A probability density function can be given as 1x on the interval -1- b- and outside this interval the value of function is zero- The value of b is

Given, probability density function f (x) = {1x 1 ≤ x ≤ b 0 amp; else where. also ∫_ ∞^∞ f(x)dx = 1 ⇒∫_ ∞^1 0 dx + ∫_1^b1x dx + ∫_b^∞ 0 dx ⇒ [ ln |x| ]^b_1 ...

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Standard XII

Mathematics

Standard Deviation

A probability...

Question

A probability density function can be given as $\frac{1}{x}$ on the interval [1, b] and outside this interval the value of function is zero. The value of b is

2.718

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Solution

The correct option is A 2.718
Given, probability density function
$f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{1}{x} 1 \leq x \leq b 0 & else where \end{matrix}$
also $\int_{- \infty}^{\infty} f (x) d x = 1$
$\Rightarrow \int_{- \infty}^{1} 0 d x + \int_{1}^{b} \frac{1}{x} d x + \int_{b}^{\infty} 0 d x$
$\Rightarrow {[l n | x |]}_{1}^{b} = 1$
$\Rightarrow l n b - l n 1 = 1$
$\Rightarrow l n b = 1$
$\Rightarrow b = e = 2.718$

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A probability density function can be given as 1x on the interval [1, b] and outside this interval the value of function is zero. The value of b is 2.718

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