Mercator chart
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Problem #6-1:
A bank has set a Standard that mortgage applications be processed within a certain number of days Of filing. If, out of a sample of 1000 Applications, 75 fail to meet this requirement, what is the epmo metric?
75 errors per Thousand is equivalent to 75000 epmo.
Problem #6-2:
Over the last year, 965 injections were administered at a clinic. Quality is measured by the proper amount of dosage as well as the Correct drug. In two instances, the Incorrect amount was given, and in one case, the wrong drug was given. What is the epmo metric?
DPMO = EPMO = [ (# of defects or errors) / (# of opportunities For error)*(# of units) ] x 1,000,000
EPMO = [ (3) / (2)*(965) ] x 1,000,000 = 1554
Problem #6-3:
Make a check sheet and then a Pareto diagram for the following car repair shop data:
Ticket No. | Work | Ticket No. | Work | Ticket No. | Work | ||
1 | Tires | 11 | Brakes | 21 | Transmission | ||
2 | Lube & oil | 12 | Lube & oil | 22 | Brakes | ||
3 | Tires | 13 | Battery | 23 | Transmission | ||
4 | Battery | 14 | Lube & oil | 24 | Brakes | ||
5 | Lube & oil | 15 | Lube & oil | 25 | Lube & oil | ||
6 | Lube & oil | 16 | Tires | 26 | Battery | ||
7 | Lube & oil | 17 | Lube & oil | 27 | Lube & oil | ||
8 | Brakes | 18 | Brakes | 28 | Battery | ||
9 | Lube & oil | 19 | Tires | 29 | Brakes | ||
10 | Tires | 20 | Brakes | 30 | Tires |
Check Sheet | |
Work Type | Frequency |
Lube And Oil | 11 |
Brakes | 7 |
Tires | 6 |
Battery | 4 |
Transmission | 2 |
Total | 30 |
Problem #6-4:
An air-conditioning repair Department manager has compiled data on the primary reason for 41 service calls For the previous week, as shown in the table below. Using the data, make a check sheet for the problem type and then construct a Pareto diagram for each type of customer.
Job Number | Problem/Cust Type | Job Number | Problem/Cust Type | Job Number | Problem/Cust Type | ||
301 | F/R | 315 | F/C | 329 | O/C | ||
302 | O/R | 316 | O/C | 330 | N/R | ||
303 | N/C | 317 | W/C | 331 | N/R | ||
304 | N/R | 318 | N/R | 332 | W/R | ||
305 | W/C | 319 | O/C | 333 | O/R | ||
306 | N/R | 320 | F/R | 334 | O/C | ||
307 | F/R | 321 | F/R | 335 | N/R | ||
308 | N/C | 322 | O/R | 336 | W/R | ||
309 | W/R | 323 | F/R | 337 | O/C | ||
310 | N/R | 324 | N/C | 338 | O/R | ||
311 | N/R | 325 | F/R | 339 | F/R | ||
312 | F/C | 326 | O/R | 340 | N/R | ||
313 | N/R | 327 | W/C | 341 | O/C | ||
314 | W/C | 328 | O/C |
Key:
Problem Type: Customer Type:
N = Noisy C = Commercial Customer
F = Equipment Failure R = Residential Customer
W = Runs Warm
O = Odor
Checksheet | |||||
Problem | |||||
Customer Type | Noisy | Failed | Odor | Warm | Totals |
Residential | 10 | 7 | 5 | 3 | 25 |
Commercial | 3 | 2 | 7 | 4 | 16 |
Totals | 13 | 9 | 12 | 7 | 41 |
Residential Customers Commercial Customers
Problem #6-5:
Suppose that a table lamp Fails to light when turned on. Prepare a Simple cause-and-effect diagram to analyze possible causes.
Problem #6-6:
The Baldrige application Summaries are excellent sources of information to learn about best Practices. Categories 4 and 7 provide Good examples of the types of measures that leading companies use. You might also want to compare measures used By small versus large companies, manufacturing versus service, and differences With not-for-profit education and health care sectors.
Problem #6-7:
Thirty samples of size 3 resulted in an Overall mean of 16.51 and average range of 1.30. Compute control limits for `x- and R-charts.
For x-bar Chart:
UCL = 16.51 + 1.02(1.30) = 17.840
LCL = 16.51 – 1.02(1.30) = 15.180
For R-chart:
UCL = 2.574(1.3) = 3.3462
LCL = 0(1.3) = 0
Problem #6-8:
Use the sample data below to construct `x- and R-charts. Assume That the sample size is 5:
Sample | x-bar | R | Sample | x-bar | R |
1 | 95.72 | 1.0 | 11 | 95.80 | 0.6 |
2 | 95.24 | 0.9 | 12 | 95.22 | 0.2 |
3 | 95.18 | 0.8 | 13 | 95.56 | 1.3 |
4 | 95.44 | 0.4 | 14 | 95.22 | 0.5 |
5 | 95.46 | 0.5 | 15 | 95.04 | 0.8 |
6 | 95.32 | 1.1 | 16 | 95.72 | 1.1 |
7 | 95.40 | 0.9 | 17 | 94.82 | 0.6 |
8 | 95.44 | 0.3 | 18 | 95.46 | 0.5 |
9 | 95.08 | 0.2 | 19 | 95.60 | 0.4 |
10 | 95.50 | 0.6 | 20 | 95.74 | 0.6 |
For n = 5, D_{4} = 2.114, D_{3} = 0, A_{2} = 0.577
95.398=(every x-bar)/20 and 0.665=(every R)/(20)
For x-bar chart:
UCL = 95.398 + 0.577(0.665) = 95.782
LCL = 95.398 - 0.577(0.665) = 95.014
For R-chart:
UCL = 2.114(0.665) = 1.406
LCL = 0
Problem #6-9:
Checkout time at a supermarket is Monitored using a mean and a range chart. Six samples of n=20 observations have been obtained and the sample means And ranges computed. Determine upper and Lower control limits for mean and range charts. Is this process in control?
Sample | Mean | Range | ||||||||
1 | 3.06 | 0.42 | ||||||||
2 | 3.15 | 0.50 | ||||||||
3 | 3.11 | 0.41 | ||||||||
4 | 3.13 | 0.46 | ||||||||
5 | 3.06 | 0.46 | ||||||||
6 | 3.09 | 0.45 | ||||||||
n = 20 | ||||||||||
A_{2} = 0.180 | = 3.10 Mean Chart: ± A_{2} = 3.10 ± 0.180(0.450) | |||||||||
D_{3}= 0.415 | = 0.450 | = 3.10 ± .081 | ||||||||
D_{4} = 1.585 | Hence, UCL is 3.181 | |||||||||
and LCL is 3.019. All means are within these limits. | ||||||||||
Range Chart: UCL is D_{4}= 1.585(0.45) = 0.7133 | ||||||||||
LCL is D_{3}= 0.415(0.45) = 0.1868 | ||||||||||
In control since all Points are within these limits. | ||||||||||
Problem #6-10:
A medical facility does MRIs for sports Injuries. Occasionally a test yields Inconclusive results and must be repeated. Using the following sample data and n=200, determine the upper and lower Control limits. Is the process in Control?
SAMPLE | ||||||||||||||||||||||||||||
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | ||||||||||||||||
Number of retests | 1 | 2 | 2 | 0 | 2 | 1 | 2 | 0 | 2 | 7 | 3 | 2 | 1 | |||||||||||||||
SAMPLE | ||||||||||||||||||||||||||||
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | ||||||||||||||||
Number of retests | 1 | 2 | 2 | 0 | 2 | 1 | 2 | 0 | 2 | 7 | 3 | 2 | 1 | |||||||||||||||
Fraction defective | 0.005 | 0.010 | 0.010 | 0.000 | 0.010 | 0.005 | 0.010 | 0.000 | 0.010 | 0.035 | 0.015 | 0.010 | 0.005 | |||||||||||||||
n = 200 control Limits: Thus, UCL is .030 and LCL becomes 0. The process is Out of control. | ||||||||||||||||||||||||||||
Problem #6-11:
The postmaster of a small western town Receives a certain number of complaints each day about mail delivery. Determine control limits using the following Data. Is the process in control?
DAY | ||||||||||||||
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | |
No. Of complaints | 4 | 10 | 14 | 8 | 9 | 6 | 5 | 12 | 13 | 7 | 6 | 4 | 2 | 10 |
C= control Limits:
UCL Is 16.266, LCL becomes 0.
All Values are within the limits so the process is in control.
Problem #6-12:
Given the following data for the number of Defects per spool of cable, is the process in control?
OBSERVATION | ||||||||||||||
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | |
No. Of defects | 2 | 3 | 1 | 0 | 1 | 3 | 2 | 0 | 2 | 1 | 3 | 1 | 2 | 0 |
C=
Control limits:
UCL is 5.17, LCL becomes 0.
All Values are within the limits so the process is in control.
Problem #6-13:
After a number of complaints about its Directory assistance, a telephone company examined samples of calls to Determine the frequency of wrong numbers give to callers. Each sample consisted of 100 calls. Determine control limits – is the process in Control?
SAMPLE | ||||||||||||||||
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
No. Of errors | 5 | 3 | 5 | 7 | 4 | 6 | 8 | 4 | 5 | 9 | 3 | 4 | 5 | 6 | 6 | 7 |
Problem #6-14:
Consider The following data showing the number of errors per thousand lines of code for A software development project. Construct A c-chart and interpret the results.
Sample 1 2 3 4 5 6 7 8 9 10
# of Errors 4 15 13 20 17 22 26 17 20 22
Center line = c-bar = 17.6
UCL = 17.6 + 3 sqrt(17.6) = 30.186
LCL = 17.6 – 3 sqrt(17.6) = 5.014
The first point is below the lower control limit, so the process Is out of control.
Problem #6-15:
Determine if these three processes are Capable:
Process Mean Standard Deviation LSL USL
1 7.5 0.10 7.0 8.0
2 4.6 0.12 4.3 4.9
3 6.0 0.14 5.5 6.7
Process 1: Cp = 8.0 – 7.0 = 1.67 (capable)
6(0.10)
Process 2: Cp = 4.9 – 4.3 = 0.83 (not Capable)
6(0.12)
Process 3: Cpu = 6.7 – 6.0 = 1.67
3(0.14)
Cpl = 6.0 – 5.5 = 1.19
3(0.14)
Cpk = (min Cpu and Cpl) = 1.19, so process is not capable
Problem #6-16:
Each Process below is non-centered with respect to the specifications for that Process. Compute the appropriate Capability index for each and decide if the process is capable.
Process | Mean | Std Deviation | Lower Spec | Upper Spec |
H | 15.0 | 0.32 | 14.1 | 16.0 |
K | 33.0 | 1.00 | 30.0 | 36.5 |
T | 18.5 | 0.40 | 16.5 | 20.1 |
Problem #6-17:
As Part of an insurance company’s training program, participants learn how to Conduct an analysis of clients’ insurability. The goal is to have participants Achieve a time in the range of 30 to 45 minutes. Test results for three participants Were: Armand, a mean of 38 minutes and a Standard deviation of 3 minutes; Jerry, a mean of 37 minutes and a standard Deviation of 2.5 minutes; and Melissa, a Mean of 37.5 minutes and a standard deviation of 1.8 minutes. Which of the participants would you judge to Be capable?
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
= Process mean, s = Process standard deviation.
LSL= 30 minutes, USL = 45 minutes,
_{Armand} = 38 minutes, s_{ Armand} = 3 minutes
_{Jerry} = 37 minutes, s_{ Jerry} = 2.5 minutes
_{Melissa} = 37.5 minutes, s_{ Melissa} = 1.8 minutes
For Armand:
Since .78 < 1.33, Armand is Not capable.
For Jerry:
Since .93 < 1.33, Jerry is Not capable.
For Melissa, since USL-X=X-LSL=7.5, the process is centered, therefore we will use C_{p} To measure process capability.
Since 1.39 > 1.33, Melissa is capable.