Please answer #2. Extra Info: Upload speeds are measured on a standard scale in

Question

Please answer #2.

Extra Info: 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 approx.. normally distributed, with a mean of 1.005 and a std. deviat. Of .10. Each day one upload speed is measured . The upload speed is considered acceptable if the measurement on the std. scale is between .95 and 1.05.

A Word document report summarizing your data findings and conclusions.

An Excel spreadsheet that supports your report. You may use PHStat or Excel.

1. Assuming that the distribution has not changed from what it is 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 reduces the standard deviation of the upload speed to .075? Explain.

Explanation / Answer

Normal Distribution
Mean ( u ) =1.005
Standard Deviation ( sd )=0.1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

a)
P(X < 1) = (1-1.005)/0.1
= -0.005/0.1= -0.05
= P ( Z <-0.05) From Standard Normal Table
= 0.4801                  

b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.95) = (0.95-1.005)/0.1
= -0.055/0.1 = -0.55
= P ( Z <-0.55) From Standard Normal Table
= 0.29116
P(X < 1) = (1-1.005)/0.1
= -0.005/0.1 = -0.05
= P ( Z <-0.05) From Standard Normal Table
= 0.48006
P(0.95 < X < 1) = 0.48006-0.29116 = 0.1889                  

c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1) = (1-1.005)/0.1
= -0.005/0.1 = -0.05
= P ( Z <-0.05) From Standard Normal Table
= 0.48006
P(X < 1.05) = (1.05-1.005)/0.1
= 0.045/0.1 = 0.45
= P ( Z <0.45) From Standard Normal Table
= 0.67364
P(1 < X < 1.05) = 0.67364-0.48006 = 0.1936                  

d)
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 0.95) = (0.95-1.005)/0.1
= -0.055/0.1= -0.55
= P ( Z <-0.55) From Standard Normal Table
= 0.2912
P(X > 1.05) = (1.05-1.005)/0.1
= 0.045/0.1 = 0.45
= P ( Z >0.45) From Standard Normal Table
= 0.3264
P( X < 0.95 OR X > 1.05) = 0.2912+0.3264 = 0.617515                  

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