# What are the mean and standard deviation of the probability density function given by (p(x))/k=x/(1-x^2 for  x in [2,4], in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Jul 17, 2016

$k = \frac{1}{-} 2.2939$

#### Explanation:

PDF means that we need to find the expectation which becomes
$E \left(x\right) = {\int}_{2}^{4} p \left(x\right) \times x$ $\mathrm{dx}$

and
$p \left(x\right) = k \left(\frac{x}{1 - {x}^{2}}\right)$

${\int}_{2}^{4} k \left(\frac{x}{1 - {x}^{2}}\right) \times x$ $\mathrm{dx}$

$k {\int}_{2}^{4} {x}^{2} / \left(1 - {x}^{2}\right) \mathrm{dx}$

$k \left[\frac{\ln \left(4 + 1\right) - \ln \left(4 - 1\right) - 2 \cdot 4}{2} - \frac{\ln \left(2 + 1\right) - \ln \left(2 - 1\right) - 2 \cdot 2}{2}\right]$

$k \left(\ln \frac{5}{2} - \ln \left(3\right) - 2\right)$

$k \left(- 2.2939\right)$
for the cummulative value to be equal to 1 then

$- 2.2939 k = 1$
then solve for k
$k = \frac{1}{-} 2.2939$