The time X that a technician requires to perform preventative maintenance on an air-conditioning unit has a mean of 1 hour and a standard deviation of 1.
a) Your company operates 70 of these units. What is the probability (proportion of customers) having to wait 50 minutes or more?
Answer
Given that mean= 1 hour = 60 minutes and standard deviation = 1 hour = 60 minutes
we have find P(X>50)
using the formula P(X>x) = P(z>(x-mean)/(s/sqrt(n)))
setting x = 50, mean = 60, s = 60 and n = 70
we get
= P(z>(50-60)/(60/sqrt(70)))
= P(z>(-10)/(7.17))
= P(z>-1.39)
= 0.9177 (using z table for -1.3 in left most column and 0.09 in top row, then selected the intersecting cell)
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