Question

Let X be a random variable with the following probability distribution. Then E(X) =

Answer 1

The expected value of a distribution(when the distribution is discrete) is given by:

`E\lbrack X\rbrack=\sum ^k_(n\mathop=1)x_np_n(x_n)`

Since we have only two values, k=2.

`E\lbrack X\rbrack=x_1p_1+x_2p_2`

From the table, we get the following information:

`\begin{gathered} x_1=10 \\ x_2=30 \\ p_1=0.1 \\ p_2=0.3 \end{gathered}`

Then, our expected value is:

`E\lbrack X\rbrack=10*0.1+30*0.3`

Then, solving this calculation we have:

`E\lbrack X\rbrack=10`