Concept explainers
a)
To determine: The fraction defective in each sample.
Introduction: Quality is a measure of excellence or a state of being free from deficiencies, defects and important variations. It is obtained by consistent and strict commitment to certain standards to attain uniformity of a product to satisfy consumers’ requirement.
a)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Calculation of fraction defective in each sample:
n | 200 | |||
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Prop defective | 0.02 | 0.01 | 0.025 | 0.045 |
Excel Worksheet:
The proportion defective is calculated by dividing the number of errors with the number of samples. For sample 1, the number of errors 4 is divided by 200 which give 0.02 as prop defective.
Hence, the fraction defective is shown in Table 1.
b)
To determine: The estimation for fraction defective when true fraction defective for the process is unknown.
Introduction: Quality is a measure of excellence or a state of being free from deficiencies, defects and important variations. It is obtained by consistent and strict commitment to certain standards to attain uniformity of a product to satisfy consumers’ requirement.
b)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Calculation of fraction defective:
The fraction defective is calculated when true fraction defective is unknown.
Total number of defective is calculated by adding the number of errors, (4+2+5+9) which accounts to 20
The fraction defective is calculated by dividing total number of defective with total number of observation which is 20 is divided with the product of 4 and 200 which is 0.025.
Hence, the fraction defective is 0.025.
c)
To determine: The estimate of mean and standard deviation of the sampling distribution of fraction defective for samples for the size.
Introduction:
Control chart:
It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.
c)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Estimate of mean and standard deviation of the sampling distribution:
Mean = 0.025 (from equation (1))
The estimate for mean is shown in equation (1) and standard deviation is calculated by substituting the value which yields 0.011.
Hence, estimate of mean and standard deviation of the sampling distribution is 0.025 and 0.011.
d)
To determine: The control limits that would give an alpha risk of 0.03 for the process.
Introduction:
Control chart:
It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.
d)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Control limits that would give an alpha risk of 0.03 for the process:
0.015 is in each tail and using z-factor table, value that corresponds to 0.5000 – 0.0150 is 0.4850 which is z = 2.17.
The UCL is calculated by adding 0.025 with the product of 2.17 and 0.011 which gives 0.0489 and LCL is calculated by subtracting 0.025 with the product of 2.17 and 0.011 which yields 0.0011.
Hence, the control limits that would give an alpha risk of 0.03 for the process are 0.0489 and 0.0011.
e)
To determine: The alpha risks that control limits 0.47 and 0.003 will provide.
Introduction:
Control chart:
It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.
e)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Alpha risks that control limits 0.47 and 0.003 will provide:
The following equation z value can be calculated,
From z factor table, the probability value which corresponds to z = 2.00 is 0.4772, on each tail,
0.0228 is observed on each tail and doubling the value gives 0.0456 which is the alpha risk.
Hence, alpha risks that control limits 0.47 and 0.003 will provide is 0.0456
f)
To determine: Whether the process is in control when using 0.047 and 0.003.
Introduction:
Control chart:
It is a graph used to analyze the process change over a time period. A control chart has an upper control limit, and lower control which are used plot the time order.
f)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Calculation of fraction defective in each sample:
n | 200 | |||
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Prop defective | 0.02 | 0.01 | 0.025 | 0.045 |
UCL = 0.047 & LCL = 0.003
Graph:
A graph is plotted using UCL, LCL and prop defective values which show that all the sample points are well within the control limits which makes the process to be in control.
Hence, the process is within control for the limits 0.047 & 0.003.
g)
To determine: The mean and standard deviation of the sampling distribution.
Introduction:
Control chart:
It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.
g)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Long run fraction defective of the process is 0.02
Calculation of mean and standard deviation of the sampling distribution:
Fraction defective in each sample:
n | 200 | |||
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Prop defective | 0.02 | 0.01 | 0.025 | 0.045 |
The mean is calculated by taking average for the proportion defective,
The values of the proportion defective are added and divided by 4 which give 0.02.
The standard deviation is calculated using the formula,
The standard deviation is calculated by substituting the values in the above formula and taking square root for the resultant value which yields 0.099.
Hence, mean and standard deviation of the sampling distribution is 0.02&0.0099.
h)
To construct: A control chart using two sigma control limits and check whether the process is in control.
Introduction:
Control chart:
It is a graph used to analyze the process change over a time period. A control chart has a upper control limit, and lower control which are used plot the time order.
h)
Answer to Problem 5P
Explanation of Solution
Given information:
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Fraction defective in each sample:
n | 200 | |||
Sample | 1 | 2 | 3 | 4 |
Number with errors | 4 | 2 | 5 | 9 |
Prop defective | 0.02 | 0.01 | 0.025 | 0.045 |
Calculation of control limits:
The control limits are calculated using the above formula and substituting the values and taking square root gives the control limits of the UCL and LCL which are 0.0398 and 0.0002 respectively.
Graph:
A graph is plotted using the fraction defective, UCL and LCL values which shows that one sample points is beyond the control region which makes the process to be out of control.
Hence, control chart is constructed using two-sigma control limits and the chart shows that the process is not in control.
Want to see more full solutions like this?
Chapter 10 Solutions
Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
- Solve the following Question 1. How do volume and variety affect the process selection and layout types? Discuss 2. How is the human resource aspect important to operation function? Discuss 3. Discuss the supply network design and its impact on the overall performance of the organization.arrow_forwardHelp with question?arrow_forwardWhat are some good examples of bullet points on a resume for a Christian Elementary School?arrow_forward
- What is an example of a cover letter for a Christian School Long-Term Substitute Teaching position?arrow_forwardThe supply chain is a conventional notion, but organizations are only really interested in making products that they can sell to customers. Provided they have reliable supplies of materials and reasonable transport for finished products, logistics is irrelevant. Do you think this is true? If yes, explain, and if no, clearly explain as well.arrow_forwardworking as a program operations managerarrow_forward
- 12 X1, X230 1 x =0x2 write the Following linear Programming model by 1- general Form Canonical Forms Canonical formY 2- Standard Form Max Z=35X+ 4 X 2 +6 X3 ST. X+2X2-5x3 = 40 3X, + 6X2 + 7x 3 = 30 7x, +lox2 x3 = 50 X3 X 2 X 3 <0arrow_forwarda/ a Minimum cost assign each worker for one job at Jobs J1 12 33 WI 2 4 6 W2 5 W3 5 33 6 7arrow_forwardوبة واضافة هذه القيمة الى القيم Ex: Assign each job for each worker at minimum total Cost عمل لكل عامل وبأقل كلفة ممكنة obs الأعمال Workors العمال J1 J2 J3 J4 W₁ 15 13 14 12 W2 11 12 15 13 W3 13 12 10 11 W4 15 17 14 16arrow_forward
- The average completion time (flow time) for the sequence developed using the FCFS rule = 11.75 days (round your response to two decimal places). The percentage utilization for the sequence developed using the FCFS rule = 42.55 % (enter your response as a percentage rounded to two decimal places). b) Using the SPT (shortest processing time) decision rule for sequencing the jobs, the order is (to resolve a tie, use the order in which the jobs were received): An Alabama lumberyard has four jobs on order, as shown in the following table. Today is day 205 on the yard's schedule. In what sequence would the jobs be ranked according to the decision rules on the left: Job Due Date A 212 B 209 C 208 D 210 Duration (days) 6 3 3 8 Sequence 1 Job B 2 3 4 A D The average tardiness (job lateness) for the sequence developed using the SPT rule = 5.00 days (round your response to two decimal places). The average completion time (flow time) for the sequence developed using the SPT rule = 10.25 days…arrow_forwardWith the aid of examples, fully discuss any five (5) political tactics used in organisations.arrow_forwarda. With the aid of examples, define discrimination. b. Fully discuss any four (4) types of discrimination in the workplacearrow_forward
- Practical Management ScienceOperations ManagementISBN:9781337406659Author:WINSTON, Wayne L.Publisher:Cengage,Operations ManagementOperations ManagementISBN:9781259667473Author:William J StevensonPublisher:McGraw-Hill EducationOperations and Supply Chain Management (Mcgraw-hi...Operations ManagementISBN:9781259666100Author:F. Robert Jacobs, Richard B ChasePublisher:McGraw-Hill Education
- Purchasing and Supply Chain ManagementOperations ManagementISBN:9781285869681Author:Robert M. Monczka, Robert B. Handfield, Larry C. Giunipero, James L. PattersonPublisher:Cengage LearningProduction and Operations Analysis, Seventh Editi...Operations ManagementISBN:9781478623069Author:Steven Nahmias, Tava Lennon OlsenPublisher:Waveland Press, Inc.