Answer to Question #303869 in Statistics and Probability for qwerty

Question

Find the mean of the probability distribution of the random variable x, which can take only the values 1, 2, and 3. Given that P(1) = 10/33, P(2) = 1/33, and P(3) = 12/33

Answer 1

Check

$$\sum _ip(x_i)=10/33+1/3+12/33=1$$

$$mean=E(X)=\sum_ix_ip(x_i)$$

$$=1(10/33)+2(1/3)+3(12/33)=58/33$$

The mean of the probability distribution of the random variable $x$ is $58/33.$