Let X be an r.v. with p.d.f. f(x) = 3x^2, for 0 ? x ? 1. (i) Calculate the EX and the median of X
and compare them. (ii) Determine the 0.125-quantile of X.
Solution
i)E(x)=integral(xf(X)dx)=integral(3x^3dx)=.75x^410=.75
median
The median of the distribution of an r.v. X is usually defined as a point denoted by XO.50, for
which
P(X <= XO.50) >=0.50 and P(X>= XO.50)>= 0.50,
P(X<=m)=integral (f(x) dx)=x^2m0=m^2=0.5=>m=.707
P(x>=m)=integral(f(x)dx)=x^21m=1-m^2=0.5=>m=.707
so median =.707
mean>median
ii)quartile is median of (0-.125)
P(x>=m)=.125=>m^2=.125
=>m=.354--.125 quartile of X