The administrator of a small town received a certain number of complaints during the last two weeks. What are the three sigma control limits for the number of complaints received each day using the following data? Is the process in control? Day = 1 2 3 4 5 6 7 8 Number of complaints = 6 11 15 10 4 9 3 20 UCL = 21.253, LCL = 1.562 and in control. UCL = 22.118, LCL = 2.371 and in control. UCL = 22.118, LCL = 2.371 and NOT in control. UCL = 19.118, LCL = 0.383 and in control. UCL = 19.118, LCL = 0.383 and NOT in control.

Respuesta :

Solution:

Calculate mean of the sample data as shown below :

x (bar) = [tex]\frac{6 + 11 + 15 + 10+ 4+ 9 +3+ 20}{8}[/tex]

          = 60.5

Calculate the upper and lower control limits as below :

UCL = x (bar) + z[tex]\sqrt{x(bar)}[/tex]

       = 60.5 + 3 [tex]\sqrt{60.5}[/tex]

      =  60.5 + 3 ( 7.778 )

      = 83.834

LCL = x (bar) - z[tex]\sqrt{x(bar)}[/tex]

      = 60.5 - 3 ( 7.778 )

      = -37.166

The three-sigma control units for complaints that are being received every day employing data would be as follows:

Mean - [tex]60.5[/tex]

Upper Control Limit - [tex]83.834[/tex]

Lower Control Limit - [tex]-37.166[/tex]

To determine the three-sigma control units,,

We will find out the mean,

Mean = (Sum of all the observations)/(No. of observations)

= 60.5

Upper Control Limit = Mean + [tex]z\sqrt{xbar}[/tex]

[tex]= 60.5 +[/tex] [tex]3\sqrt{60.5}[/tex]

[tex]= 60.5 + 3(7.778)[/tex]

[tex]= 83.834[/tex]

Lower Control Limit = [tex]Mean -[/tex] [tex]z\sqrt{xbar}[/tex]

= [tex]60.5 - 3(7.778)[/tex]

= [tex]-37.166[/tex]

Learn more about 'Mean' here:

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