Question

MANAGING ASHLAND MULTI-COMM SERVICES

The AMS technical services department has embarked on a quality improvement effort. It’s first project relates to maintaining the target upload speed for its Internet service subscribers. Upload speeds are measured on a standard scale in which the target value is 1.0. Data collected over the past year indicate that the upload speed is approximately normally distributed, with a mean of 1.005 and a standard deviation of 0.10. Each day, one upload speed is measured. The upload speed is considered acceptable if the measurement on the standard scale is between 0.95 and 1.05

1. Assuming that the distribution has not changed from what it was in the past year, what is the probability that the upload speed at any time is:

a. Less than 1.0?

b. Between 0.95 and 1.0?

c. Between 1.0 and 1.05?

d. Less than 0.95 or greater than 1.05?

2. The objective of the operations team is to reduce the probability that the upload speed is below 1.0. Should the team focus on process improvement that increases the mean upload speed to 1.05, or on process improvement that reduces the standard deviation of the upload speed to 0.75? Explain.

Answer 1

    X be the upload speed                          
   X follows normal distribution mean μ and standard deviation σ                          
   Given    μ = 1.005       σ = 0.10              
   We use Excel function NORM.DIST to find the probabilities                          
                              
1a)   To find P(X < 1.0)                          
   P(X < 1.0) = NORM.DIST(1.0, 1.005, 0.1, TRUE)                          
       = 0.4801                      
   P(X < 1.0) = 0.4801                        
                              
b)   To find P(0.95 < X < 1.0)                          
   P(0.95 < X < 1.0) = P(X < 1.0) - P(X < 0.95)                          
       = NORM.DIST(1.0, 1.005, 0.1, TRUE) - NORM.DIST(0.95, 1.005, 0.1, TRUE)                      
       = 0.4801 - 0.2912                      
       = 0.1889                      
   P(0.95 < X < 1.0) = 0.1889                          
                              
c)   To find P(1.0 < X < 1.05)                          
   P(1.05 < X < 1.0) = P(X < 1.05) - P(X < 1.0)                          
       = NORM.DIST(1.05, 1.005, 0.1, TRUE) - NORM.DIST(1.0, 1.005, 0.1, TRUE)                      
       = 0.1936                      
   P(1.0 < X < 1.05) = 0.1936                          
                              
d)   To find P(X < 0.95 OR X > 1.05)                          
   P(X < 0.95 OR X > 1.05) = 1 - P(0.95 < X < 1.05)                          
       = 1 - [NORM.DIST(1.05, 1.005, 0.1, TRUE) - NORM.DIST(0.95, 1.005, 0.1, TRUE)]                      
       = 1 - 0.3825                      
       = 0.6175                      
   P(X < 0.95 OR X > 1.05) = 0.6175                        
                              
                              
2)   It is definitely better to have a process improvement which                          
   reduces the standard deviation of the upload speed to 0.0075 provides a better                           
   reduction in probability that the upload speed is below 1.0                          
   Probability is majorly affected if there is a change in scale than when there is a change in origin.