Answer to Question #88410 in Physics for Shivam Nishad

Question

A continuous random variable X has the p.d.f.
f(x)={ kx³(1-x), 0≤x≤1
{ 0 , otherwise
Calculate the mean and standard deviation of X .

Answer 1

The probability density function

$$f(x)=\left\{\begin{matrix} k x^3(1-x), & 0\leq x\leq1 \\ 0 & \rm{otherwise} \end{matrix}\right.$$

$$1=\int_{-\infty}^{\infty}f(x)dx=\int_{0}^{1}k x^3(1-x)dx=\frac{k}{20}$$

So

$$k=20$$

$$f(x)=\left\{\begin{matrix} 20 x^3(1-x), & 0\leq x\leq 1 \\ 0 & \rm{otherwise} \end{matrix}\right.$$

The mean value

$$\mu=\overline{x}=\int_{-\infty}^{\infty}xf(x)dx=\int_{0}^{1}20 x^4(1-x)dx=\frac{2}{3}$$

Since

$$\overline{x^2}=\int_{-\infty}^{\infty}x^2f(x)dx=\int_{0}^{1}20 x^5(1-x)dx=\frac{10}{21}$$

we obtain that standard deviation

$$\sigma=\sqrt{\overline{x^2}-(\overline{x})^2}=\sqrt{\frac{10}{21}-\left(\frac{2}{3}\right)^2}=\frac{1}{3}\sqrt{\frac{2}{7}}=0.178$$