Question

Customers experiencing technical difficulty with their internet cable hookup may call an 800 number for technical support. It takes the technician between 30 seconds and 10 minutes to solve the problem. The distribution of this support time follows the uniform distribution.a. What are the values for a and b in minutesb. What is the mean time to resolve the problem? 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 percent of the problem solving times. What are the end points of these two times?

Answer 1

Answer:

a) a= 0.5 minutes

b = 10 minutes

b) Mean= 5.25 minutes

Standard deviation = 2.74

c) The percentage that the problem.

takes more than 5 minutes is.

52.63%

d) The end points are 2.875 and.

7.625

Step-by-step explanation:

The number of technical supports that may be called is 800

It takes the technician between 30 seconds to 10 minutes to resolve a problem.

a) Since the distribution of the technical support is uniform,

a = 30 seconds

a = (30/60)minutes

a = 0.5 minutes

b = 10 minutes

b) The mean to resolve the problem is U

U = (a + b) /2

= (0.5 + 10)/2

= 10.5/2

= 5.25 minutes

The standard deviation of the time is α

α = √(b-a)^2/12

α = √(10 -0.5)^2/12

= √9.5^2/12

= √90.25/12

= √7.5208

= 2.74

c) To find the percentage that it takes more than 5 minutes to resolve the problem note that if it takes more than 5 minutes to solve a problem, the time taken is between 5 minutes and 10 minutes. We have

(b - x) / (b - a)

Where x = 5

= 10 - 5/ 10 - 0.5

= 5/9.5

= 0.5263

= 0.5263*100

= 52.63%

d) Suppose the probability of solving a problem is at the middle (50%),

0.5/2 = (b-x) / (b-a)

0.25 = (10-x) / (10-0.5)

0.25 = (10-x)/9.5

10-x = 0.25*9.5

10-x = 2.375

x= 10- 2.375

x = 7.625 (this is for the upper side)

For the lower side, we have 0.5+2.375

= 2.875

Therefore, the range is 2.875 minutes to 7.625 minutes

Answer 2

Using the uniform distribution, we have that:

a) $a = 0.5, b = 10$.

b) The mean of these times is of 4.75 minutes, with a standard deviation of 2.74 minutes.

c) 52.63% of the problems take more than 5 minutes to resolve.

d) The endpoints of these two times are 2.875 minutes and 7.625 minutes.

An uniform distribution has two bounds, a and b.

The probability of finding a value of at lower than x is:

$P(X < x) = \frac{x - a}{b - a}$  

The probability of finding a value above x is:

$P(X > x) = \frac{b - x}{b - a}$

Item a:

• Uniformly distributed between 30 seconds and 10 minutes, thus, in minutes, $a = 0.5, b = 10$.

Item b:

The mean of the uniform distribution is:

$M = \frac{b - a}{2}$

Then

$M = \frac{10 - 0.5}{2} = 4.75$

The standard deviation of the uniform distribution is:

$S = \sqrt{\frac{(b - a)^2}{12}}$

Then

$S = \sqrt{\frac{(10 - 0.5)^2}{12}} = 2.74$

The mean of these times is of 4.75 minutes, with a standard deviation of 2.74 minutes.

Item c:

$P(X > 5) = \frac{10 - 5}{10 - 0.5} = 0.5263$

0.5263 = 52.63% of the problems take more than 5 minutes to resolve.

Item d:

• The uniform distribution is symmetric, which means that the middle 50% is between the 25th percentile and the 75th percentile.
• The 25th percentile is X for which P(X < x) = 0.25.
• The 75th percentile is X for which P(X < x) = 0.75.

25th percentile:

$0.25 = \frac{x - 0.5}{10 - 0.5}$

$x - 0.5 = 0.25(9.5)$

$x = 2.875$

75th percentile:

$0.75 = \frac{x - 0.5}{10 - 0.5}$

$x - 0.5 = 0.75(9.5)$

$x = 7.625$

The endpoints of these two times are 2.875 minutes and 7.625 minutes.

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