Q1 (Joint probability). We know X|Y is a normal random variable with mean Y and variance 2. The probability distribution of Y is a binomial distribution with success probability 0.3 and the number of trials 5. What is the expected value of X?
We can find out the expected value of Y initially using the formula E(Y) = n*p = 5*0.3 = 1.5
Variance of Y = npq = 5*0.3*0.7 =1.05
E(X/Y) = 2
Variance of X = npq = 1.05
Expected value of X = E(X) = n*p = 1.5
Q1 (Joint probability). We know X|Y is a normal random variable with mean Y and variance...
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