PLEASE SHOW EXCEL STEPS SAMS has embarked on a quality improvement effort. Its f

Question

PLEASE SHOW EXCEL STEPS

SAMS has embarked on a quality improvement effort. Its 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 values are 1.0. Data collected over the past year indicated that the upload speed is approximately normaly distributed, with a mean of 1005 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 4. If the distribution has not changed from what it was in the past year, what is the probability that the upload speed is a. Less than 1.0? b. Between 0.95 and 1.0? . Less than 0.95 or greater than 1.05 The objective of the operations team is to reduce the probability that the upload speed is below 10. Should the team focus on process improvement that increases the mean upload speed to 105 or on process improvement that reduces the standard deviation of the upload speed to 0075? S. Continuing the quality improvement effort, the upload speed for SAMS internet service subscribers has been monitored. As before, 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 speeds are approximately normally distributed, with a mean of 1.005 and a standard deviation of 0.10 6. Each day, at 25 random times, the upload speed is measured. If the distribution has not changed from what it was in the past year, what is the probability that the upload speed is a. Less than 1.0? b. Between 0.95 and 1.0? c. Less than 0.95 or greater than 1.05 7. Compare your results in question #4 and question #6. what condusions can you reach concerning the differences?

Explanation / Answer

Sol #4:

mean = 1.005

standard deviation = 0.10

a)probability that upload speed is Less than 1.0 i.e P(X<1.0)

z value of 1 is (1-1.005)/0.10 = -0.005/0.10 = -0.05 corresponding p value is 0.48

P(X<1.0) = 0.48

b)

z value of 0.95 is (0.95-1.005)/0.10 = -0.55 corresponding p value is 0.29116

P(X<0.95) = 0.29116

P(0.95<X<1) = 0.48 - 0.29116 = 0.18884

c)

z value of 1.05 is (1.05-1.005)/0.10 = 0.45 corresponding p value is 0.29116

P(X<1.05) = 0.673645

P(X<0.95 or X>1.05) = P(X<0.95)+P(X>1.05) = P(X<0.95)+1 -P(X<1.05)=0.29116+1 -0.673645 = 0.617515

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