Question

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 1

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)