The time (in hours) that a technician requires to perform preventive maintenance on an air-conditioning unit is very different from unit to unit. Some problems take hours to fix, some can be done in minutes. Probability says that individual 'time' is governed by an exponential distribution, the precise nature of which is not an issue in Math 1530, except that we can use the mean time ,u = 1 hour and the standard deviation 0'= 1 hour from that individual-units distribution. Your company has a contract to maintain 70 of these units in an apartment building. While you do not know exactly how much time each unit will take, you must schedule technicians' time for a visit to this building based on the average time x for the 70 units. Is it safe to budget an average of 1.1 hours per unit? Or should you budget an average of 1.25 hours? We believe that the manufacturing and distribution process associated with this type of air-conditioning unit is such that variation from one unit to the next is random. Thus, we treat these 70 air conditioners as an SR5 from all units of this type. The central limit theorem says that the sample mean time x spent working on 70 units has approximately the Normal distribution with mean equal to the population mean [1 = 1 hour and standard deviation (TA/7O = 0.12 hour. The 'sampling distribution' of x is therefore approximately N(1, 0.12). Question 1 (1 point) 1. Which graph represents the sampling distribution of R ? O O Question 2 (1 point) Use this distribution to find the probability that the average maintenance time for 70 units )'( is less than 1 hour. 0 - 50% Q 68% Q 50% Q o Question 3 (1 point) Use this distribution and the 68-95-99] Rule to find the probability that the average maintenance time )'( for 70 units is between 0.76 and 1.24 hours. Q 95% O 99.7% 0 68% 0 cannot be determined without an actual sample. Question 4 (1 point) Use this distribution and the 68-95-997 Rule to find the probability that the average maintenance time R for 70 units is less than 0.88 hours. 0 84% Q 68% Q 34% Q 16% Question 5 (1 point) You need to use this distribution to find the probability that the average maintenance time )'( for 70 units exceeds 1.1 hours. For that, you must first compute the z-score for )'( = 1.1 in the sampling distribution. The value of the z- score is Question 6 (1 point) Find the probability that the average time x exceeds 1.1 hours? Oz = 0.83 0.2033 0.7967 Oz = 0.17Question 7 (1 point) What is the probability that the average time exceeds 1.25 hours? 0.0188 0.9812 2.08 - 2.08 O 2.08 O 0.9812 0 0.0188 ecall that the individual air conditioning units ("air conditioners") have repair times that can be wildly different from one individual unit to the next. We've also been given information that the population of individual 'repair times' has mean time u = 1 hour and the standard deviation 0'= 1 hour. A new contractor in a different city works with a complex that includes 100 of these air conditioning units. We can still treat these 100 air conditioners as an 5R5 (size n = 100) from all units of this type. Question 8 (1 point) The central limit theorem says that the sample mean time i spent working on 100 units has approximately the Normal distribution with mean equal to the population mean u = 1 hour and standard deviation O'/\\ 3 O 0.9 O 0.1 Q 1 Q 0.12 Question 9 (1 point) Use this new distribution to find the probability that the average maintenance time for 100 units X is less than 0.88 hours. 0 - 1.2 Q 0.8849 0 16% 0 0.1151 Question 10 (1 point) Use this new distribution to find the probability that the average maintenance time for 100 units R is greater than 1.1 hours. Q 0.4062 0 0.8413 Q1 Q 0.1587