SOLUTION: 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 d
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X = number of d
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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? Answer by oscargut(2103) (Show Source):
You can put this solution on YOUR website! 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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