Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will serve a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?

The time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will serve a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?

According to the given details, the average time of servicing is $1$hour, the standard deviation is 1 hour and the sample size is $70$units. To compute the probability average time of servicing is $1.1$hours per technician to complete the work, use the Ti-$83$calculator, for this, click on $2$^nd, then DISTR and then scroll down to the normal CDF option and enter the furnished details. After this, click on ENTER button of the calculator to have the desired outcome. The screenshot is given as below:

Accordingly, compute probability average time of servicing is $1.1$hours per technician to finish the work is approximately $0.8$. Thus, it means that $80$% chance that the service time to do the work will be less than$1.1$ hours.