BADM3601 HW #5
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School
George Washington University *
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Course
3601
Subject
Philosophy
Date
Feb 20, 2024
Type
Pages
4
Uploaded by ProfSteelCaribou34
Justine Phi
Homework #5
A.
R-Chart and X-Bar Chart (20 Points):
a.
Compute the upper and lower control limits for the sample means and
ranges of the thermostat testing process.
For sample size (n)=5, the factors from table 5.1 three-sigma limits control chart
are
= 0.577 ,
= 0 and
= 2.115.
𝐴
2
𝐷
3
𝐷
4
Control Limits of Mean Chart:
UCL = X_double bar +
* R-bar
𝐴
2
= 71.928 + 0.577 * 2.54 = 73.39358
73.394
≈
LCL = X_double bar -
* R-bar
𝐴
2
= 71.928 - 0.577 * 2.54 = 70.46242
70.462
≈
Control Limits of Range Chart:
UCL =
* R-bar
𝐷
4
= 2.115 * 2.54 = 5.3721
5.372
≈
LCL =
* R-bar
𝐷
3
= 0 * 2.54 = 0
b.
Draw an R-chart and x- bar chart for this process. Examine both charts. Is
this process in control? Explain.
Justine Phi 2
The sample means and ranges are within their LCL and UCL ranges therefore, the process
variability is under statistical control.
B.
P-Chart (10 Points):
The process is in control because all the proportions lie within the control limit.
C.
C-Chart (10 Points):
a.
Assuming 10 observations are adequate for this purpose, determine the
three-sigma control limits for dimples per camper top.
Justine Phi 3
# of samples (n) = 10
Total # of dimples (c) = 7 + 9 + 14 + 11 + 3 + 12 + 8 + 4 + 7 + 6 = 81
Mean = 81/10 = 8.1
UCL = c-bar +3 * (c-bar)^0.5 = 8.1 + 3 * (8.1) ^ 0.5 =16.6381
LCL = c-bar - 3 * (c-bar)^0.5 = 8.1 - 3 * (8.1) ^ 0.5 = 0 (can’t be negative)
All the values in the sample lie within the control limits, therefore, the sample is in
control.
b.
Suppose that the next camper top has 15 dimples. What can you say about
the process now?
If the next camper top has 15 dimples, the number of dimples are still within the
upper and lower control limits range therefore, the process is still in control.
D.
Line Balancing (10 Points):
a.
What is the theoretical minimum number of stations?
b.
How many stations are required using the longest work element decision
rule?
Applying the longest work element decision rule, 6 stations are required.
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Justine Phi 4
c.
Suppose that a solution requiring five stations is obtained. What is its
efficiency?
Efficiency = (274/(5*60))*100 = 91.3333333 %