Question

A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the likelihood that people were absent 0, 1, 2, 3, 4, or 5 days last month.

Number of days absent 0 1 2 3 4 5   

Probability 0.5 0.2 0.15 0.1 0.04   0.01

Is this a probability distribution? Why? What is the random variable? What is the probability that a randomly selected employee was absent for more than 3 days? What is the probability that a randomly selected employee was absent for at least 3 days?

Answer 1

The total sum of all the individual probabilities = 0.5 + 0.2 + 0.15 + 0.1 + 0.04 + 0.01 = 1

Since the sum of all the individual probabilities is equal to 1, so it is a probability distribution.

The random variable is the number of days was absent by the employees in the last month.

P(X > 3) = P(X = 4) + P(X = 5)

              = 0.04 + 0.01 = 0.05

P(X > 3) = P(X = 3) + P(X = 4) + P(X = 5)

              = 0.1 + 0.04 + 0.01

              = 0.15