2 value: 1.66 points Customers experiencing technical difficulty with their inte

Question

2 value: 1.66 points Customers experiencing technical difficulty with their internet cable hookup may call an 800 number for technical support. It takes the technician between 90 seconds and 15 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? (Do not round your intermediate calculations. Round your answers to 1 decimal place.) 1.5 b-1. What is the mean time to resolve the problem? (Do not round your intermediate calculations Round your answer to 2 decimal places.) Mean 8.25 b-2. What is the standard deviation of the time? (Do not round your intermediate calculations. Round your answer to 2 decimal places.) Standard deviation 3.89 c. What percent of the problems take more than 5 minutes to resolve? (Do not round your intermediate calculations. Round your answer to 2 decimal places.) Percent 74.07 % d. Suppose we wish to find the middle 50% of the problem-solving times. What are the end points of these two times? (Do not round your intermediate calculations. Round your answers to 3 decimal places.) End point1 End point 2 11.63

Explanation / Answer

a)

Values of a and b in minutes :

a = 90/ 60 = 1.5 minutes

b = 15 minutes

b-1)

Mean time to resolve the problem = ( b + a ) / 2 = 16.5 / 2 = 8.25 minutes

b-2)

Standard deviation of the time =( ( b -a )2 /12 )0.5 = 3.89

c)

Percent of problems take more than 5 minutes to resolve = P (X > 5)

We know that P(X < x) = ( x -a ) / (b - a)

Hence,  P (X > 5) = 1 - P (X < 5) = 1 - ( x -a ) / (b - a) = 1 - ( 5 - 1.5 ) / (15 - 1.5 ) = 1 - 0.259 = 0.7407 = 74.07 %

d)

For middle 50 % values we need to obtain first quartile q1 and third quartile q3

P( X < q1 ) = 0.25

( q1 - 1.5 ) / ( 15 -1.5 ) = 0.25

q1 = 4.87

P( X < q3 ) = 0.75

( q3 - 1.5 ) / ( 15 -1.5 ) = 0.75

q3 = 11.63

We have,

End point 1 = 4.87

End point 2 = 11.63

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