Can somebody show me how to find the median and variance of this piece wise function? I've been trying so hard to do it.
f(x) =
4/(x^5) , x >= 1
0, otherwise
Ans: 1.11, 7/9
Hello,
Compute the cdf :
F(t)=0 if t<1
$\displaystyle F(t)=\int_{1}^t 4x^{-5}~dx=1-\frac{1}{t^4}$ if t>1
The median is the value m such that $\displaystyle F(m)=0.5$. Shouldn't be too difficult...
The variance is $\displaystyle E[X^2]-E[X]^2=\int_1^\infty x^2f(x) ~dx-\left(\int_1^\infty xf(x) ~dx\right)^2=\dots$
Note : the fact that it is piecewise is not that important. Here, it just tells you the range of the distribution.