I need help with this question Please provide a very detaile

I need help with this question. Please provide a very detailed response and I will give you the most points allowable. The most detailed answer will get the points. Take your time and please be correct. Lets see how good you are!....Thanks!!

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.

Solution

mean = 1.005 and a standard deviation = .10

Z = (xbar - mu)/(stdev/sqrt(n))

a) probability that the upload speed is less than 1.0 = P[Z < (1 - 1.005)/(0.10/sqrt(1))]

                                                                                     = P[Z < -0.05] = NORMSDIST(-0.05)

                                                                                     = 0.4801

b) probability that the upload speed is between .95 and 1.0

= P[Z < (1 - 1.005)/(0.10/sqrt(1))] - P[Z < (0.95 - 1.005)/(0.10/sqrt(1))]

= P[Z < -0.05] - P[Z < -0.55]

= NORMSDIST(-0.05) - NORMSDIST(-0.55)

= 0.48 - 0.2911

= 0.1889

c) probability that the upload speed is between 1.0 and 1.05

= P[Z < (1.05 - 1.005)/(0.10/sqrt(1))] - P[Z < (1 - 1.005)/(0.10/sqrt(1))]

= P[Z < 0.45] - P[Z < -0.05]

= NORMSDIST(0.45) - NORMSDIST(-0.05)

= 0.6736 - 0.4801

= 0.1935

d) probability that the upload speed is less than .95 or greater than 1.05

= P[Z < (0.95 - 1.005)/(0.10/sqrt(1))] + 1 - P[Z < (1.05 - 1.005)/(0.10/sqrt(1))]

= P[Z < -0.55] + 1 - P[Z < 0.45]

= NORMSDIST(-0.55) + 1 - NORMSDIST(0.45)

= 0.2911 + 1 - 0.6736

= 0.6175

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2) If the std dev is reduced upto .075

probability that the speed is less than 1 is = P[Z < (1 - 1.005)/(0.075/sqrt(1))]

                                                                     = P[Z < -0.067]

                                                                     = 0.4734

If mean is 1.05

probability that the speed is less than 1 is = P[Z < (1 - 1.05)/(0.10/sqrt(1))]

                                                                     = P[Z < -0.5]

                                                                     = 0.3085

Hence the team should focus on process improvement that increases the mean upload speed to 1.05 as the probability of upload speed below 1.0 is lesser in this case