Information The time (in hours) that a technician requires to perform preventive
ID: 3127534 • Letter: I
Question
Information
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 = 1 hour and the standard deviation = 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 SRS 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 hour and standard deviation /70 0.12 hour. The ‘sampling distribution’ of x is therefore approximately N(1, 0.12).
1.) Use this distribution to find the probability that the average maintenance time for 70 units x is less than 1 hour.
2.) Use this distribution and the 68-95-99.7 Rule to find the probability that the average maintenance time x for 70 units is between 0.76 and 1.24 hours.
3.) Use this distribution and the 68-95-99.7 Rule to find the probability that the average maintenance time for 70 units x is less than 0.88 hours.
4.) You need to use this distribution to find the probability that the average maintenance time x for 70 units exceeds 1.1 hours. For that, you must first compute the z-score for x = 1.1 in the sampling distribution. The value of the z-score is
5.) Find the probability that the average time x exceeds 1.1 hours
6.) What is the probability that the average time exceeds 1.25 hours?
7.) 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 SRS (size n = 100) from all units of this type. The central limit theorem says that the sample mean time x spent working on 100 units has approximately the Normal distribution with mean equal to the population mean = 1 hour and standard deviation /n
8.) Use this new distribution to find the probability that the average maintenance time for 100 units x is less than 0.88 hours.
9.) Use this new distribution to find the probability that the average maintenance time for 100 units x is greater than 1.1 hours.
Explanation / Answer
1)
As 1 hour is the mean, then half of the sample means should be less than it,
P(x<1 hr) = 0.5 [ANSWER]
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2)
0.76 and 1.24 hrs are 2 standard errors away from the mean, so
ANSWER: 95% [ANSWER]
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3)
0.88 is 1 standard deviation below the mean, so
P(x<0.88) = (100 - 68)/2 = 16% [ANSWER]
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4)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 1.1
u = mean = 1
s = standard deviation = 0.12
Thus,
z = (x - u) / s = 0.833333333 [ANSWER]
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5)
Thus, using a table/technology, the right tailed area of this is
P(z > 0.833333333 ) = 0.202328381 [ANSWER]
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