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Let x1 and x2 be independent random variables with mean m and varian.pdf

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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.

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- 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

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