A teacher is monitoring how often students visit the website of the course during the day. She finds the following probability d
istribution. Find the expected number of visits to the course website.
visits(x) 0 1 2 3
probability 0.45 0.35 0.15 0.05
0.75
0.25
0.80
0.85
1 answer:
Answer:
Option C: 0.80
Step-by-step explanation:
Expected number of visits, E(x) = ∑xp(x)
∑ = Summation
x = visits
p(x) = probability of a number of visits
E(x) = (0*0.45) + (1*0.35) + (2*0.15) + (3*0.05)
E(x) = 0 + 0.35 + 0.30 + 0.15 = 0.80
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