Dear Student, Mean x =EX =∑i=13xi pxi =x1 px1+x2 px2+x3 px3 =1 ×12+2 ×13+3 ×16 =12+13+16 =3+2+16=66=1 EX2 =∑i=13xi2 pxi =x12 px1+x22 px2+x32 px3 =12 ×12+22 ×13+32 ×16 =12+43+96 =3+8+96=206=103Thus, variance is given by, varX=EX2-EX2 =103-12 =103-1 =73 Regards

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No links please The probability distribution f(x) variable X is gven by, x p(x) 1 1 ä 2 1 of a discrete random 3 1 6 Find Mean and Variance of X.

Dear Student,

Mean (x) = E (X) = \sum_{i = 1}^{3} x_{i} p (x_{i}) = x_{1} p (x_{1}) + x_{2} p (x_{2}) + x_{3} p (x_{3}) = 1 \times \frac{1}{2} + 2 \times \frac{1}{3} + 3 \times \frac{1}{6} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1 E (X^{2}) = \sum_{i = 1}^{3} {x_{i}}^{2} p (x_{i}) = {x_{1}}^{2} p (x_{1}) + {x_{2}}^{2} p (x_{2}) + {x_{3}}^{2} p (x_{3}) = {(1)}^{2} \times \frac{1}{2} + {(2)}^{2} \times \frac{1}{3} + {(3)}^{2} \times \frac{1}{6} = \frac{1}{2} + \frac{4}{3} + \frac{9}{6} = \frac{3 + 8 + 9}{6} = \frac{20}{6} = \frac{10}{3} Thus, variance is given by, var (X) = E (X^{2}) - {(E (X))}^{2} = \frac{10}{3} - {(1)}^{2} = \frac{10}{3} - 1 = \frac{7}{3}

Regards

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No links please No links please The probability distribution f(x) variable X is gven by, x p(x) 1 1 ä 2 1 of a discrete random 3 1 6 Find Mean and Variance of X.

No links please

No links please The probability distribution f(x) variable X is gven by, x p(x) 1 1 ä 2 1 of a discrete random 3 1 6 Find Mean and Variance of X.