Question

Customers experiencing technical difficulty with their Internet cable hookup may call an 800 number for technical support. Itakes the technician between 30 seconds and 12 minutes to resolve the problem. The distribution of this support time follows the uniform distribution.
A) What percent of the problems takes more than 5 minutes to resolve? (Do not round your intermediate calculations. Round your answer to 2 decimal places.)
B) Suppose we wish to find the middle 50% of the problem-solving times. What are the end points of these two times? (Donot round your intermediate calculations. Round your answers to 3 decimal places.)
C) What are the values for a and b in minutes? (Do not round your intermediate calculationsRound your answer to 1 decimal place.)(D-1) What is the mean time to resolve the problem? (Do not round your intermediate calculations. Round your answer to 2 decimal places.)
(D-2) What is the standard deviation of the time? (Do not round your intermediate calculations. Round your answer to 2 decimal place

Answer 1

Answer:

(A) The percent of the problems takes more than 5 minutes to resolve is 60.87%.

(B) The problem solving time will be 50% as long as the difference between the two end points is 5.75.

(C) a = 0.50 minutes, b = 12 minutes.

(D) Mean = 6.25 minutes, Standard deviation = 3.32 minutes

Step-by-step explanation:

Let the random variable X represent the time it takes the technician to resolve the problem.

The random variable X follows an Uniform distribution with parameters a =  0.50 minutes and b = 12 minutes.

The probability density function of X is:

$f_{X}(x)=\frac{1}{b-a};\ a$

(A)

Compute the probability that the problems takes more than 5 minutes to resolve as follows:

$P(X>5)=\int\limits^{12}_{5}{\frac{1}{12-0.50}}\, dx$

               $=\frac{1}{11.50}\cdot \int\limits^{12}_{5}{1}\, dx \\\\=\frac{1}{11.50}\cdot [x]\limits^{12}_{5}\\\\=\frac{1}{11.50}\cdot [12-5]\\\\=\frac{7}{11.50}\\\\=0.608696\\\\\approx 0.6087$

Thus, the percent of the problems takes more than 5 minutes to resolve is 60.87%.

(B)

Let the middle 50% of the problem-solving times be between u and v.

Then,

P (u < X < v) = 0.50

$\int\limits^{v}_{u}{\frac{1}{12-0.50}}\, dx=0.50\\\\\frac{1}{11.50}\cdot \int\limits^{v}_{u}{1}\, dx=0.50\\\\\frac{v-u}{11.50}=0.50\\\\(v-u)=5.75$

Thus, the problem solving time will be 50% as long as the difference between the two end points is 5.75.

(C)

The interval in which the technician can solve the problem is 30 seconds to 12 minutes.

So, the values of a and b in minutes is:

a = 30 seconds = 0.50 minutes

b = 12 minutes.

(D)

The mean time is:

$\mu=\frac{a+b}{2}=\frac{0.50+12}{2}=6.25\ \text{minutes}$

The standard deviation of the time is:

$\sigma=\sqrt{\frac{(b-a)^{2}}{12}}=\sqrt{\frac{(12-0.50)^{2}}{12}}=3.3197\approx 3.32\ \text{minutes}$