Question 773061
Suppose the following table represents the probability distribution for the number of days that an employee is absent from work during a week selected at random.

X = number of days absent

X              0                1              2
probability    .80            .15             .05

a.) what is the probability that an employee could be absent for more than 2 days?


B.) what is the expected value for the number of  days absent?

c.)what is the variance for the number of days absent

d.) what is the standard deviation fro the number of days absent?

Answers:

a) 0

b) E(X)= 0(0.8)+1(0.15)+2(0.05) = 0.25

c) V(X) = E(X^2)-(E(X))^2 

E(X^2) =(0^2)(0.8)+(1^2)(0.15)+(2^2)(0.05) = 0.15+0.2 = 0.35

Then V(X) = 0.35- (0.25)^2 = 0.35-0.0625 = 0.2875

d) sd(X) = sqrt(0.2875) 

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