Question

Suppose that a random variable x can take on integer values from 0 to 5 and its pdf is defined as follows:P(x)=11-2x/36 Calculate the expected values of x

Answer 1

Given the pdf defined:

`P(X=x)=(11-2x)/(36)`

The formula to find the expected value of x for a discrete distribution is

`E(X)=\sum ^(\infty)_(n\mathop=-\infty)xP(X=x)`

Here, x ranges from 0 to 5. Find E(X).

`\begin{gathered} E(X)=\sum ^5_(n\mathop=0)xP(X=x) \\ =0\cdot P(x=0)+1\cdot P(x=1)+2\cdot P(x=2)+3\cdot P(x=3)+4.P(x=4)+5\cdot P(x=5)P( \\ =0+1\cdot(11-2\cdot1)/(36)+2\cdot(11-2\cdot2)/(36)+3\cdot(11-2\cdot3)/(36)+4\cdot(11-2\cdot4)/(36)+5\cdot(11-2\cdot5)/(36) \\ =(9)/(36)+(14)/(36)+(15)/(36)+(12)/(36)+(5)/(36) \\ =(55)/(36) \end{gathered}`

which is the expected value of x.