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A teacher is monitoring how often students visit the website of the course during the day. She finds the following probability distribution. 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

Respuesta :

ukshedrack

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

the expected number of visits to the course website Option C: 0.80

Calculation of expected no of visits:

We know that

Expected number of visits, E(x) = ∑xp(x)

Here,

∑ = Summation

x = visits

p(x) = probability of a number of visits

So,

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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