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?

Question

Quinn Allen · Accepted Answer

#3.3361599515537#

if the PDF of #x# is #f(x)=1/x# on the interval of #[e,e^2]# then #int_e^(e^2) f(x) dx= 1#

The standard deviation expected is provided by

#E[sigma]=sqrt(int_e^(e^2) (x-mu)^2 f(x)dx )#

#=sqrt(int_e^(e^2) (x^2-2xmu+mu^2)/xdx)#

the integral being #x^2/2 -2mux+mu^2ln(x) #

and #E[sigma] =sqrt( (e^4-e^2)/2 + mu^2 + 2mu - 2mu^2)#

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

#E[mu]=int_e^(e^2) xf(x)dx #

#=int_e^(e^2) 1/x*x dx #

#=e^2-e = 4.6707742704716#

#E[sigma] = 3.3361599515537#

Ezekiel Anderson · Answer

undefined To find the standard deviation of a continuous random variable with a probability density function $ f(x) = \frac{1}{x} $ on the interval $ [e, e^2] $, we first need to find the mean, denoted by $ \mu $, and then use it to calculate the standard deviation.

The mean $ \mu $ of a continuous random variable is given by the integral of $ x $ times the probability density function $ f(x) $ over the interval $ [e, e^2] $, divided by the integral of $ f(x) $ over the same interval:

$$ \mu = \frac{\int_{e}^{e^2} x \cdot \frac{1}{x} dx}{\int_{e}^{e^2} \frac{1}{x} dx} $$

$$ = \frac{\int_{e}^{e^2} dx}{\int_{e}^{e^2} \frac{1}{x} dx} $$

$$ = \frac{e^2 - e}{\ln(e^2) - \ln(e)} $$

$$ = \frac{e^2 - e}{2\ln(e) - \ln(e)} $$

$$ = \frac{e^2 - e}{2} $$

Now, the standard deviation $ \sigma $ can be calculated using the formula:

$$ \sigma = \sqrt{\int_{e}^{e^2} (x - \mu)^2 \cdot \frac{1}{x} dx} $$

$$ = \sqrt{\int_{e}^{e^2} \left(x - \frac{e^2 - e}{2}\right)^2 \cdot \frac{1}{x} dx} $$

This integral will need to be evaluated, but first, we need to simplify the expression inside the integral by expanding $ (x - \frac{e^2 - e}{2})^2 $. Once we simplify and integrate, we can then take the square root to find $ \sigma $.