Answer to Question #329454 in Statistics and Probability for Hassam

Question

 Let X be a discrete random variable with the following PMF: pX(x) = 1/2 for x = 0 1/3 for x = 1 1/6 for x = 2 0 otherwise (i) Compute SX, the range of the random variable X. (ii) Calculate P(X ≤ 0.5). (iii) Calculate P(0.25 < X < 0.75). (iv) Calculate P(X = 0.2|X < 0.6).

Answer 1

$i:\\EX=0\cdot \frac{1}{2}+1\cdot \frac{1}{3}+2\cdot \frac{1}{6}=\frac{2}{3}\\EX^2=0^2\cdot \frac{1}{2}+1^2\cdot \frac{1}{3}+2^2\cdot \frac{1}{6}=1\\varX=EX^2-\left( EX \right) ^2=1-\left( \frac{2}{3} \right) ^2=\frac{5}{9}\\sX=\sqrt{varX}=\sqrt{\frac{5}{9}}=0.745356\\range\left( X \right) =2-0=2\\ii:\\P\left( X\leqslant 0.5 \right) =P\left( X=0 \right) =\frac{1}{2}\\iii:\\P\left( 0.25<X<0.75 \right) =P\left( X\in \emptyset \right) =0\\iv:\\P\left( X=0.2|X<0.6 \right) =\frac{P\left( X=0.2 \right)}{P\left( X=0 \right)}=\frac{0}{1/2}=0$