AP EAMCET201823 Apr 2018Morning ShiftMathematicsProbabilityActual
A random variable X has the following probability distribution The variance of this random variable is
Options
- A0
- B5 24
- C3 24
- D7 4
Correct answer
D. 7 4
Step-by-step solution
We have, We know that sum of probability =1 . aligned & p (X=x_i )=1 & array l 1 6 +k+ 1 4 +k+ 1 6 =1 2 k=1- 7 12 k= 5 24 array & Now, E(X)= x_i p_i & =-2 ( 1 6 )-1 ( 5 24 )+0 ( 1 4 )+1 ( 5 24 )+2 ( 1 6 ) & array c E(x)=0 Now, E (X^2 )= x_i^2 p i =(-2)^2 ( 1 6 )+(-1)^2 ( 5 24 )+0 ( 1 4 ) = 4 6 + 5 24 + 5 24 + 4 6 = 16+10+16 24 = 42 24 = 7 4 array & array l var (x)=E(x)^2-[E(x)]^2 var (x)= 7 4 -0= 7 4 array aligned