If random variable X has a probability density function of f(x)=1/x on the interval[e,e^2], what is the standard deviation of X?

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If random variable X has a probability density function of f(x)=1/x on the interval[e,e^2], what is the standard deviation of X?

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Answer 1 · 2016-07-22T14:43:06

$3.3361599515537$ $3.3361599515537$

Explanation:

if the PDF of $x$ $x$ is $f (x) = \frac{1}{x}$ $f(x)=1/x$ on the interval of $[e, e^{2}]$ $[e,e^2]$ then $\int_{e}^{e^{2}} f (x) d x = 1$ $int_e^(e^2) f(x) dx= 1$

The expected standard deviation is given by

$E [σ] = \sqrt{\int_{e}^{e^{2}} {(x - μ)}^{2} f (x) d x}$ $E[sigma]=sqrt(int_e^(e^2) (x-mu)^2 f(x)dx )$

$= \sqrt{\int_{e}^{e^{2}} \frac{x^{2} - 2 x μ + μ^{2}}{x} d x}$ $=sqrt(int_e^(e^2) (x^2-2xmu+mu^2)/xdx)$

the integral being
$\frac{x^{2}}{2} - 2 μ x + μ^{2} ln (x)$ $x^2/2 -2mux+mu^2ln(x)$

and
$E [σ] = \sqrt{\frac{e^{4} - e^{2}}{2} + μ^{2} + 2 μ - 2 μ^{2}}$ $E[sigma] =sqrt( (e^4-e^2)/2 + mu^2 + 2mu - 2mu^2)$

now we need to solve for $μ$ $mu$ and we shall have our final answer

$E [μ] = \int_{e}^{e^{2}} x f (x) d x$ $E[mu]=int_e^(e^2) xf(x)dx$

$= \int_{e}^{e^{2}} \frac{1}{x} \cdot x d x$ $=int_e^(e^2) 1/x*x dx$

$= e^{2} - e = 4.6707742704716$ $=e^2-e = 4.6707742704716$

$E [σ] = 3.3361599515537$ $E[sigma] = 3.3361599515537$

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If random variable X has a probability density function of f(x)=1/x on the interval[e,e^2], what is the standard deviation of X?

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