# What are the mean and standard deviation of the probability density function given by (p(x))/k=ln(x^3+1) for  x in [0,1], in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Nov 30, 2017

The mean is $= 0.78$ and the standard deviation is $= 0.18$

#### Explanation:

The function $\ln \left({x}^{3} + 1\right)$ is continuous and positive on the interval $\left[0 , 1\right]$, so it's a probability density function.

First , determine $k$

$P \left(x\right) = k \ln \left({x}^{3} + 1\right)$

${\int}_{0}^{1} p \left(x\right) \mathrm{dx} = k {\int}_{0}^{1} \ln \left({x}^{3} + 1\right) \mathrm{dx} = 1$

$k \times 0.2 = 1$

$k = \frac{1}{0.2} = 5$

Therefore,

$P \left(x\right) = 5 \ln \left({x}^{3} + 1\right)$

So,

The mean is

$E \left(x\right) = \int x P \left(x\right) \mathrm{dx} = 5 {\int}_{0}^{1} x \ln \left({x}^{3} + 1\right) = 0.78$

The variance is

$V a r \left(x\right) = E \left({x}^{2}\right) - {\left(E \left(x\right)\right)}^{2}$

$E \left({x}^{2}\right) = \int {x}^{2} P \left(x\right) \mathrm{dx} = {\int}_{0}^{1} 5 {x}^{2} \ln \left({x}^{3} + 1\right) \mathrm{dx} = 0.64$

Therefore,

$V a r \left(x\right) = 0.64 - {0.78}^{2} = 0.0316$

The standard deviation is

$\sigma \left(x\right) = \sqrt{V a r \left(x\right)} = \sqrt{0.0316} = 0.18$