Final answer:
Although the question seems to be confused about UST service technician certification intervals, the answer focuses on determining if budgeting 1.1 hours to service air conditioning units is enough. Using the central limit theorem and standard deviation, the calculations suggest that while often 1.1 hours may be sufficient, there is approximately a 20% chance that this time budget won't be adequate to cover the service time variability.
Step-by-step explanation:
A UST service technician must be recertified by the equipment manufacturer at the time interval recommended by the equipment manufacturer or at least every 24 months according to the Environmental Protection Agency (EPA) guidelines for UST service technicians. However, the question provided seems to be more related to operations management and how to allocate time for technicians to service air conditioning units, which falls under the realm of Business or Operations Management rather than UST service technician certification intervals.
To determine whether budgeting an average of 1.1 hours per technician to service each unit is sufficient, we need to consider the average service time and the variability in service times. Given that the average time to service a unit is one hour with a standard deviation of one hour, we can use the central limit theorem to find out if 1.1 hours will be likely enough for a sample of 70 units.
By using the central limit theorem, we know that the sampling distribution of the sample mean will be approximately normally distributed because we have a large enough sample size (n=70). The standard error of the mean (SEM) is the standard deviation divided by the square root of the sample size. SEM = 1 / sqrt(70) ≈ 0.119. Now we can calculate the z-score for 1.1 hours to see how likely it is that the service time will exceed this time limit.
Z = (X - μ) / SEM where X is the average time budgeted (1.1 hours), μ is the population mean (1 hour). Plugging in the values: Z = (1.1 - 1) / 0.119 ≈ 0.84. Looking up a z-score of 0.84 on the z-table finds that roughly 80% of the sample means will be less than 1.1 hours, indicating there is a 20% chance that the average service time could exceed the budgeted 1.1 hours.
Therefore, while 1.1 hours may often be sufficient, there is a chance it would not accommodate the variability adequately, considering the standard deviation of one hour and the fact that the z-score outcome signals that there is some risk involved. If the company wants to be more conservative in their planning, they might consider budgeting a bit more time per technician per unit.