Question

Customers experiencing technical difficulty with their Internet cable hookup may call an 800 number for technical support. It takes the technician between 28 seconds to 20 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? (Round your intermediate & final answers to 2 decimal places.) (b-1) What is the mean time to resolve the problem? (Round your intermediate & final answer to 2 decimal places.) (b-2) What is the standard deviation of the time? (Round your intermediate & final answer to 2 decimal places.) (c) What percent of the problems take more than 5 minutes to resolve? (Round your intermediate & final answer to 2 decimal places.) (d) Suppose we wish to find the middle 50% of the problem-solving times. What are the end points of these two times? (Round your intermediate values to 2 decimal places & final answers to 3 decimal places.)

Answer 1

Solution :

Given that, it takes the technician between 28 seconds to 20 minutes to resolve the problem.

a) So the value of a and b in minutes.

a = 28/60 = 0.47 minute and b = 20 minute

b-1) Mean

b-2) Standard deviation

c) Percent of problems take more than 5 minues

We know that for uniform distribution , P(X <x) = (x-a)/(b-a)

d) Given that , middle area is 50% = 0.5

so area between endpoints x1 and x2 is 0.5.

Now area below x1 is 0.25.

So using P( X <x1) = (x1-a)/(b-a)

=> 0.25 = (x1-0.47)/(20-0.47)

=> 0.25 = (x1-0.47)/ 19.5

=>x1-0.47 = 0.25* 19.5

=>x1 = 4.88 + 0.47 = 5.353

Similarly area below x2 is 0.75

so again using P(X < x2) =(x2-a)/(b-a)

=> 0.75 = (x2-0.47)/(20-0.47)

=> 0.75 = (x2-0.47)/ 19.5

=> x2-0.47= 0.75*19.5

=>x2-0.47 = 14.625

=>x2 = 14.6+0.47

=>x2 = 15.095

So end points are (5.353,15.095)