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Suppose that X is the number of bears spotted by tourists to Banff, Canada. The table below is the probability distribution for X. What is the expected value of X, that is, what is the mean of its distribution?

X 0 1 2 3 4

Probability 0.2 0.1 0.4 0.15 0.15

Answer :

ukshedrack

Answer:

Expected value of X = 1.95

Step-by-step explanation:

Expected value of X, E(X) = ∑XP(X)

∑ = Summation

X = number of bears spotted

P(X) = probability of a certain number of bears spotted

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*0.1) + (2*0.4) + (3*0.15) + (4*0.15)

E(X) = 0 + 0.1 + 0.8 + 0.45 + 0.60 = 1.95

The Expected value of X = 1.95

Calculation of expected value:

We know that

Expected value of X, E(X) = ∑XP(X)

here

∑ = Summation

X = number of bears spotted

P(X) = probability of a certain number of bears spotted

Now

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 × 0.1) + (2 ×  0.4) + (3 × 0.15) + (4 × 0.15)

E(X) = 0 + 0.1 + 0.8 + 0.45 + 0.60

= 1.95

Learn more about the probability here: https://brainly.com/question/20306112

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