Question

Need some assistance explaining what the answers mean, can solve the math portion, struggling to put them into words for a report.

The AMS technical services department 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 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 .10. Each day, one upload speed is measured. The upload speed is considered acceptable if the measurement on the standard scale is between .95 and 1.05.

Assuming that 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 .95 and 1.0? c.) between 1.0 and 1.05? d.) less than .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 .075? Explain.

Call attention to significant information by setting it off visually with bullets, lists or indentation • Include visual aids to emphasize and clarify the text • Use an informal style • Give each paragraph a topic sentence • Link paragraphs by using transitional words and phrases • Strive for readability by using short sentences, concrete words, and appropriate terminology • Be accurate, thorough, and impartial in presenting the material • Document for all material quoted or paraphrased from secondary sources using footnotes

Answer 1

Answer to question# 1)

The question simply tells us that the upload speed is check daily and based on that data it is observed that the mean (M) = 1.005, Standard deviation ( s) = 0.10

P(x < 1)

Z formula is:

Z = (x – M) / S

Z = (1 – 1.005) / 0.1

Z = -0.05

P(z < -0.05) = 0.4800

.

P(0.95 < x < 1.0) = P(x < 1.0) – P(x < 0.95)

P(x<1.0) = 0.4800

P(x < 0.95)

Z = (0.95-0.1005)/0.10

Z = -0.55

P(x < 0.95) = 0.29116

.

P(0.95 < x < 1.0) = 0.4800 -0.29116 = 0.18884

.

P(1 < x < 1.05) = P(x < 1.05) – P(x < 1)

P(x < 1.05)

Z = (1.05-1.005)/0.10 = 0.45

P(z< 0.45) = 0.6736

P(1 < x < 1.05) = 0.6736-0.4800 = 0.1936

.

P(x < 0.95 or x > 1.05) = 1 – P(0.95< x < 1.05)

P(0.95 < x < 1.05) = P9x < 1.05) – P(x < 0.95)

P(0.95 < x < 1.05) = 0.6736 – 0.29116 = 0.38244

.

P(x<0.95 or x > 1.05) = 1 – 0.38244 = 0.61756