Let mean and standard deviation of probability distribution ( array |r|r|c|c|c| X =x & -3 & 0 & 1 & P ( X =x) & 1 4 & ~K & 1 4 & 1 3 array ) be and respectively and if - =2 then =

Step-by-step solution

The sum of probabilities must equal 1, so K is found from 1 4 + K + 1 4 + 1 3 = 1 yielding K = 1 6 . The mean is = xP(X=x) = (-3) ( 1 4 ) + 0 ( 1 6 ) + 1 ( 1 4 ) + ( 1 3 ) = - 1 2 + 3 . The second moment is E(X^2) = 9 ( 1 4 ) + 0 + 1 ( 1 4 ) + ^2 ( 1 3 ) = 5 2 + ^2 3 . The variance becomes ^2 = E(X^2) - ^2 = 5 2 + ^2 3 - (- 1 2 + 3 )^2 = 9 4 + 2 ^2 9 + 3 . Given - = 2 , squaring gives ^2 = ( + 2)^2 = ( 3 2 + 3 )^2 = 9 4 + + ^2 9 . Equating both expressions for ^2 and simplifying leads to ^2 - 6 = 0 , so ( - 6) = 0