If the mean of the following probability distribution of a random variable X : array |c|c|c|c|c|c| X & 0 & 2 & 4 & 6 & 8 P ( X ) & a & 2 a & a+b & 2 b & 3 b array is 46 9 , then the variance of the distribution is

Step-by-step solution

aligned & P_i=1 & a+2 a+a+b+2 b+3 b=1 aligned 4 a+6 b=1 ...(I) E ( x )= mean = 46 9 aligned & P_i X_i= 46 9 4 a+4 a+4 b+12 b+24 b= 46 9 & 8 a+40 b= 46 9 aligned 4 a+20 b= 23 9 ...(II) Subtract (I) from (II) we get aligned & b = 1 9 & a = 1 12 & Variance = E ( x _ i ^2 )- E ( x _ i )^2 & E ( x _ i )^2=0^2 9^2+2^2 2 a +4^2( a + b )+6^2(2 ~b )+8^2(3 ~b ) & =24 a +280 ~b aligned Put a = 1 12 ~b = 1 9 aligned & E ( x _ i ^2 )=2+ 280 9 = 298 9 & ^2= E ( x _ i ^2 )- E ( x _ i )^2 & = 298 9 - ( 46 9 )^2 & ^2= 298 9 - 2116