# 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, D4 = 2.114, D3 = 0, A2 = 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 A2 = 0.180 = 3.10 Mean Chart: ± A2 = 3.10 ± 0.180(0.450) D3= 0.415 = 0.450 = 3.10 ± .081 D4 = 1.585 Hence, UCL is 3.181 and LCL is 3.019. All means are within these limits. Range Chart: UCL is D4= 1.585(0.45) = 0.7133 LCL is D3= 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 Cp To measure process capability.

Since 1.39 > 1.33, Melissa is capable.