# What are the mean and standard deviation of the probability density function given by p(x)=k/x  for  x in [3,8], with k being a constant such that the cumulative density is equal to 1?

Aug 25, 2016

$\mu = 5.1$
$\sigma = 1.43$

#### Explanation:

the mean is just the expected value defined as
$\int p \left(x\right) x \mathrm{dx}$
and expected standard deviation
$\sqrt{\int p \left(x\right) {\left(\mu - x\right)}^{2} \mathrm{dx}}$
finally we know that
${\int}_{3}^{8} p \left(x\right) \mathrm{dx} = 1$

first solve for $k$
now we solve for k and find out what the standard deviation and mean should be.

${\int}_{3}^{8} \frac{k}{x} \mathrm{dx} = 1$
${\int}_{3}^{8} \frac{1}{x} \mathrm{dx} = \frac{1}{k}$
${\left[\ln \left(x\right)\right]}_{3}^{8} = \frac{1}{k}$

$k = \frac{1}{\ln \left(8\right) - \ln \left(3\right)} = 1.02$

now solving for mean
${\int}_{3}^{8} k \frac{x}{x} \mathrm{dx} = k \left(8 - 3\right) = 5 k = 5.1$

now for standard deviation

$\sqrt{{\int}_{3}^{8} k \frac{{\left(x - 5.1\right)}^{2}}{x} \mathrm{dx}} = \sqrt{k {\int}_{3}^{8} x - 10.2 + \frac{26.01}{x} \mathrm{dx}}$
$= \sqrt{k \left({\int}_{3}^{8} x \mathrm{dx} - {\int}_{3}^{8} 10.2 \mathrm{dx} + {\int}_{3}^{8} \frac{26.01}{x} \mathrm{dx}\right)}$
$= \sqrt{k \left({\left[{x}^{2} / 2\right]}_{3}^{8} - {\left[10.2 x\right]}_{3}^{8} + {\left[26.01 \ln \left(x\right)\right]}_{3}^{8}\right)} = \sqrt{k \left(27.5 - 51 + 25.51\right)} = \sqrt{2.05} = 1.43$