Question 1202780: Customers experiencing technical difficulty with their Internet cable service may call an 800 number for technical support. It takes the technician between 18 seconds and 10 minutes to resolve the problem. The distribution of this support time follows the uniform distribution.
a. What are the values for a and b in minutes?
b-1. What is the mean time to resolve the problem?
b-2. What is the standard deviation of the time?
c. What percent of the problems take more than 5 minutes to resolve?
d. Suppose we wish to find the middle 50% of the problem-solving times. What are the end points of these two times?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 18 seconds is 0.3 min
a=0.3; b=10
mean is half way between the two or 10.3/2=5.15 m or 5m9s
variance is (a-b)^2/12=7.84, so sd is sqrt(V)=2.8 m or 2m48s
more than 5 minutes is (10-5)/9.7, the entire interval. That is 51.5%
The 25% percentile is 1/2 way to the median or half way between 0.3 and 5.15 or 2.725 m
The 75%ile is 10.3-2.725=7.575 m. The endpoints are [2.725m, 7.575m]
The difference between the two is 4.85 m, half the range of the the distribution, which it should be.
The 5%ile is 5% of the range, which is 5% of 9.7 m or 0.485 m
add that to the 0%ile of 0.3 m, and the answer is 0.785 m.
These can all be converted to seconds if desired.
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