Probability Distributions
- A random variable \(X\) has \(P(X=x)=kx\) for \(x=1,2,3,4\). Then \(E(X)\) is Answer: C. 3 k(1+2+3+4)=1⇒ k=frac110 ; E(X)=1+4+9+16/10=3 . 1
- The p.d.f. of \(X\) is \(f(x)=kx^2\) for \(0\le x\le3\) and 0 otherwise. Then \(P(X\le1)\) is Answer: C. frac127 ∫₀³kx²dx=9k=1⇒ k=1/9 ; P(X≤1)=1/9·1/3=frac127 . 1