# What are the mean and standard deviation of the probability density function given by (p(x))/k=x^3-4x for  x in [0,2], in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Dec 7, 2015

The mean is $\mu = \frac{16}{15} \approx 1.0667$ and the standard deviation is $\sigma = 2 \frac{\sqrt{11}}{15} \approx 0.4422$

#### Explanation:

If $p \left(x\right) = k \left({x}^{3} - 4 x\right)$ for $x \in \left[0 , 2\right]$, then ${\int}_{0}^{2} k \left({x}^{3} - 4 x\right) \mathrm{dx} = k {\left[{x}^{4} / 4 - 2 {x}^{2}\right]}_{0}^{2} = k \left(4 - 8\right) = - 4 k$. Therefore, we require $- 4 k = 1$ so that $k = - \frac{1}{4}$.

Hence, the density function is $p \left(x\right) = \frac{1}{4} \left(4 x - {x}^{3}\right) = x - {x}^{3} / 4$.

The mean is the expected value of $X$:

$\mu = E \left[X\right] = {\int}_{0}^{2} x \cdot p \left(x\right) \setminus \mathrm{dx} = {\int}_{0}^{2} \left({x}^{2} - {x}^{4} / 4\right) \setminus \mathrm{dx}$

$= {\left[{x}^{3} / 3 - {x}^{5} / 20\right]}_{0}^{2} = \frac{8}{3} - \frac{32}{20} = \frac{16}{15}$.

The expected value of ${X}^{2}$ is

$E \left[{X}^{2}\right] = {\int}_{0}^{2} {x}^{2} \cdot p \left(x\right) \setminus \mathrm{dx} = {\int}_{0}^{2} \left({x}^{3} - {x}^{5} / 4\right) \setminus \mathrm{dx}$

$= {\left[{x}^{4} / 4 - {x}^{6} / 24\right]}_{0}^{2} = 4 - \frac{64}{24} = \frac{4}{3}$.

Hence, the variance is ${\sigma}^{2} = V a r \left(X\right) = E \left[{X}^{2}\right] - {\left(E \left[X\right]\right)}^{2} = \frac{4}{3} - {\left(\frac{16}{15}\right)}^{2} = \frac{44}{225}$.

The standard deviation is

$\sigma = \sqrt{{\sigma}^{2}} = \sqrt{\frac{44}{225}} = \frac{\sqrt{44}}{15} = \frac{2 \sqrt{11}}{15}$