Question

Find the Variance of a Discrete Random Variable
For the # of H when we flip a coin twice in a discrete distribution:
Data (respectively)
X: 0,1,2
P(x): (1/4) (1/2) (1/4)
mu=1
Please help me. Thank you!

Answer 1

\(Var(X)=E(X^2)-[E(X)]^2=E(X^2)-\mu^2\)

Now, \(E(X)=\mu \)
And they already give you that value, which is 1

So, \(E(X^2)=\sum\limits_x x^2\cdot p(x)=0^2\left(\frac{1}{4} \right)+1^2\left(\frac{1}{2} \right)+2^2\left(\frac{1}{4} \right)\)