Extra Info Upload speeds are measured on a standard scale in

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.

Solution

1a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1      
u = mean =    1.005      
          
s = standard deviation =    0.1      
          
Thus,          
          
z = (x - u) / s =    -0.05      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.05   ) =    0.480061194 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.95      
x2 = upper bound =    1      
u = mean =    1.005      
          
s = standard deviation =    0.1      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.55      
z2 = upper z score = (x2 - u) / s =    -0.05      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.291159687      
P(z < z2) =    0.480061194      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.188901507   [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    1      
x2 = upper bound =    1.05      
u = mean =    1.005      
          
s = standard deviation =    0.1      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.05      
z2 = upper z score = (x2 - u) / s =    0.45      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.480061194      
P(z < z2) =    0.67364478      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.193583586   [ANSWER]

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d)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.95      
x2 = upper bound =    1.05      
u = mean =    1.005      
          
s = standard deviation =    0.1      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.55      
z2 = upper z score = (x2 - u) / s =    0.45      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.291159687      
P(z < z2) =    0.67364478      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.382485093      

Thus, those outside this interval is the complement =    0.617514907   [ANSWER]  
  

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