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Unformatted text preview: 821 a)The histogram represents a bimodal skew.3.53.02.52.01.51.00.5987654321Service TimesFrequencyMean2.090StDev0.7590N30Normal Histogram of Wally's Service Timesb)The drivethru service times appear to be in control. Even though observation 16 is high on the moving range chart, it is still within the limits of the MR chart.282522191613107414321ObservationIndividual Value_X= 2.090UCL= 4.191LCL= 0.011282522191613107412.41.81.20.60.0ObservationMoving Range__MR= 0.79UCL= 2.581LCL= 0I ndividual  MR chart for Wally's Service Timesc)The mean for this process is given in the charts as 2.090 minutes. The standard deviation is given by the equation σ = MR/d2σ = 0.79 / 1.128 = .7004 is the standard deviation for the process at Wally Burgers.d)Zupper= (4 – 2.090) / .7004 = 2.73 This z value gives us a value of .003167 in the table. This corresponds to a .317 % of the total service times that I would estimate to be over the 4 minute service time goal.e)To get our defects to stay fewer than 3.4 customers per million we need our Z value to be 4.5.To achieve this I use the equation Z = (UL – Xbar) / σ.We have the UL (upper limit), and the standard deviation so the equation gives us 4.5 = (4 – Xbar) / .7004Xbar = .8482 This tells us that for our defects to be fewer than 3.4 customers per million our average service time should be .8482 minutes when our variance of service time remains the same.f)Using the same formula as part e and changing the value of our constants, in this case we have an unknown standard deviation, we can find what the variance needs to be to have fewer than 3.4 customers per million under the 4 minute maximum time limit when we know our average time. Z = (UL – Xbar) / σ4.5 = (4 –2.090) / σσ = .4244 Given the standard deviation that we need to have, and knowing the actual measured standard deviation we can calculate how much we need to reduce the variance so that we meet our goal. σ2measured – σ2calculated= reduction of variance..70042 .42442= reduction of varianceWe need to reduce the process variance by .8002 to ensure we have fewer than 3.4 customers per million that we exceed the 4 minute service time maximum. This represents an 63.3% reduction in variance. [(.70042 .42442)/ .70042= .6328 => 63.28%]93 a)The chart shown appears to be in control. There is no reason to believe it is out of control.926a)The Xbar chart is out of control at some points. This could be caused by the plates not being perfectly symmetrical. Some plates are obviously more symmetrical than others, but they all seem to have some variability.28252219161310741256254252250SampleSample Mean__X=253.668UCL=255.216LCL=252.121282522191613107416.04.53.01.50.0SampleSample Range_R=3.202UCL=6.417LCL=01111XbarR Chart of Diameter Measurments (mm)b)The source of variation found in this section would be due to withinplate differences....
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This note was uploaded on 11/14/2011 for the course IE 300 taught by Professor Aikens during the Fall '10 term at University of Tennessee.
 Fall '10
 Aikens

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