The time T that technician requires to perform preventive maintenance on an air conditioning unit has an unknown probability distribution. However, its mean is known to be 2 hours, with standard deviation 1 hour. Suppose the company maintains 70 of these units and that the conditions of the Central Limit Theorem apply. What is the probability that a maintenance operation will take more than 2 hours and 15 minutes?

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Answer:

0.4

Step-by-step explanation:

To calculate the probability that a maintenance operation will take more than 2 hours and 15 minutes. We can first calculate the probability that ALL maintenance operation on 70 of the units will take less than 2 hours and 15 minutes, then subtract it from 1.

So the probability of a maintenance operation that would take less than 2 hours and 15 minutes, or 135 minutes is:

[tex]P(X \leq 135, \mu = 120, \sigma = 60) = 0.6 [/tex]

So the probability that a maintenance operation will take more than 2 hours and 15 minutes is:

[tex] 1 - 0.6 = 0.4[/tex]

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