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

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ukshedrack ukshedrack · Answer 1 · 2020-02-24T06:50:00+01:00

Answer:

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

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andromache andromache · Answer 2 · 2022-02-15T11:38:54+01:00

The expected value of X is Option C: E(X) = 1.65

We know that

Expected value of X, E(X) = ∑XP(X)

here

∑ = Summation

X = number of hours

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

So,

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

Learn more about the mean here: https://brainly.com/question/14634922