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Exercise 19-22 (LO19-4, LO19-6) A new machine has just been installed to cut and rough-shape large...

Exercise 19-22 (LO19-4, LO19-6)

A new machine has just been installed to cut and rough-shape large slugs. The slugs are then transferred to a precision grinder. One of the critical measurements is the outside diameter. The quality control inspector randomly selected five slugs each half-hour, measured the outside diameter, and recorded the results. The measurements (in millimeters) for the period 8:00 a.m. to 10:30 a.m. follow.

Outside Diameter (millimeters)
Time 1 2 3 4 5
8:00 87.3 87.5 88.0 87.1 87.0
8:30 86.9 88.5 87.6 87.5 87.4
9:00 87.8 88.5 86.9 87.4 88.1
9:30 86.0 88.0 87.2 87.6 87.1
10:00 86.8 86.8 86.8 87.2 87.3
10:30 88.0 86.2 87.3 87.2 87.9

  1. Determine the control limits for the mean and the range.(Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 2 decimal places.)

Mean of sample means =    LCL =    UCL =   
Mean of sample ranges =         LCL =         UCL =      

Exercise 19-30 (LO19-5, LO19-6)

Lahey Motors specializes in selling cars to buyers with a poor credit history. Listed below is the number of cars that were repossessed from Lahey customers because they did not meet the payment obligations over each of the last 36 months.

9 5 4 7 2 11 11 7 12 4
10 10 6 9 14 15 19 12 7 9
11 8 14 8 11 9 12 14 6 17
14 5 6 8 12 13

  1. Find the control limits from the c-bar data for the number repossessed. (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 2 decimal places.)

Mean number of defects per unit               
LCL
UCL

Solutions

Expert Solution

(first part)

tiem x1 x2 x3 x4 x5 mean range
8:00 87.3 87.5 88 87.1 87 87.38 1.0
8:30 86.9 88.5 87.6 87.5 87.4 87.58 1.6
9:00 87.8 88.5 86.9 87.4 88.1 87.74 1.6
9:30 86 88 87.2 87.6 87.1 87.18 2.0
10:00 86.8 86.8 86.8 87.2 87.3 86.98 0.5
10:30 88 86.2 87.3 87.2 87.9 87.32 1.8
m= 6 6
sum= 524.18 8.5
mean=sum/m= 87.36 1.42

mean of sample means==87.36, LCL=84.06, UCL=90.66

LCL=-A2*=87.36-2.326*1.42=84.06 and UCL=+A2*=87.36+2.326*1.42=90.66

for n=5, the A2=2.326

mean of sample ranges==1.42, LCL=D3=0*1.42=0, UCL=D4*=2.114*1.42=3.00

(second part)

=sum(c)/m=351/36=9.75

LCL=-sqrt()=9.75-sqrt(9.75)=6.63

UCL=+sqrt()=9.75+sqrt(9.75)=12.87

orchestra answered 2 years ago
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