Notice that the expected value of Y depends on the value of X that is observed. Tomake this clear, let’s define a new random variable Z that is equal to the expected valueof Y for a particular value of X = x.(c) [5 marks] Determine the probability distribution of Z. That is, list every possiblevalue that Z can take along with its associated probability

Given the following discrete probability distribution of the discrete random variable Z:

Let Y be a random variable as followsP(Y=-6) = 0.1P(Y=-1) = 0.1P(Y=0) = 0.1P(Y=1) = cthen what is the value of c (give your answer to 1dp)?

Z is a standard normal random variable. The P(1.20 < z < 1.85) equalsGroup of answer choices0.38490.08290.85270.4678

Find the Expected value of the following distributionx -20 -10 30 75 80P(x) 3/20 1/5 1/2 1/10 1/20(a) 20.5(b) 21.5(c) 22.5(d) 24.5

Based on the below table, answer the following question:xP(x)30.0540.4050.3060.1570.10What is the expected value?Question 1Answera.4.85b.2.48c.0.3