Solution

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Statistics for Business
Discrete Distributions

Random Variables and Probability Distributions – Activity 2 Solution
a) i) P(X > 2)

P (X > 2) = P (X = 3) + P (X = 4)
= 0.03 + 0.01
= 0.04

a) ii) P (1 ≤ X < 3)
P (1 ≤ X < 3) = P (X = 1) + P (X = 2)
= 0.08 + 0.06
= 0.14

b) What is the probability that the machine needs to be checked?

P (X > 1) = P (X = 2) + P (X = 3) + P (X = 4)
= 0.06 + 0.03 + 0.01
= 0.9

 University of Western Sydney

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