Suppose that X is the number of hours that a computer is used in a computer lab on campus. The table below is the probability distribution for X. What is the expected value of X, that is, what is the mean its distribution?X 0 1 2 3 4Probability 0.2 0.2 0.4 0.15 0.05A. 0.8 B. 1.0 C. 1.65 D. 2

Option C: E(X) = 1.65

Step-by-step explanation:

Expected value of X, E(X) = ∑XP(X)

∑ = Summation

X = number of hours

P(X) = probability of a certain number of hours

E(X) = (0*P(X=0)) + (1*P(X=1)) + (2*P(X=2)) + (3*P(X=3)) + (4*P(X=4))

E(X) = (0*0.2) + (1*0.2) + (2*0.4) + (3*0.15) + (4*0.05)

E(X) = 0 + 0.2 + 0.8 + 0.45 + 0.20 = 1.65

E(X) = 1.65 - Option C

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .