Answer to Question #298862 in Statistics and Probability for mariel

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

 Given that P(1)= 10/33 P (2) =11/33 and P(3)= 12/33, the mean is given as,

$E(x)=\sum xp(x)=(1\times {10\over33})+(2\times{11\over33})+(3\times {12\over33})={10\over33}+{22\over33}+{36\over33}={68\over33}=2.060606$

The mean of the probability distribution is 2.060606