Question 8 Why is the presence of errors that lead to scrapping bad for process
ID: 360371 • Letter: Q
Question
Question 8 Why is the presence of errors that lead to scrapping bad for process performance? C All the above. Because they might increase the variability of the process. C Because they might increase the pipeline inventory of the process E Because they might decrease the flow rate of the process. Question9 A sample average which is outside of the control limits C Is impossible to be observed. should lead to the investigation of the process. Indicates that the products in the sample are all defective. E Indicates that no process variation is present. Question 10 Suppose the production of a product has 10 sequential steps, each of which has a yield of 95%. Then the yield of the process Is lower than 95% ls 95% Is higher than 95%. C May vary t: Question 11 The output of a process is considered acceptable if it weighs between 99 and 101 grams. The process which produces this product delivers an average weight of 100 grams with a standard deviation of 1 grams. The fraction of products that are out of specs is about: 15%Explanation / Answer
Answer to question 8 :
Scrapping of products due to error requires increases process variability requiring reworking and hence can increase work in process inventory. Also , scrapping of products reduces the target output rate of sound products and thus reduces flow rate .
Therefore , correct answer would be “All of the above”
ANSWER : ALL OF THE ABOVE
Answer to Question 9:
When a sample average is outside the control limits, it unnecessarily increases process variability. But since it is an “average”, an out of limits average does not mean that all products in the sample are defective What it requires is through investigation of the process to find out root cause of such deviation. Therefore, correct answer should be” A sample average which is outside of the control limits should lead to investigation of the process”
ANSWER : A) SHOULD LEAD TO INVESTIGATION OF THE PROCESS
Answer to question 10:
When a process with 10 sequential steps with each having a yield of 95%, ( 0.95), the process yield will be calculated as = ( 0.95)^10 x 100 percent.
This value of ( 0.95)^10 will be less than 0.95.
Therefore , yield of the process is less than 95%
YILED OF THE PROCESS IS LOWER THAN 95%
Answer to question 11 :
Given are ,
Upper Specification Limit = USL = 101 gm
Lowe Specification Limit = LSL = 99 gm
Average weight = M = 100 gm
Standard deviation = Sd = 1 gm
From above , it can be seen that both USL and LSL have 1 sigma ( i.e. 1 x Sd) spread from mean ( m = 100).
Thus probability for Z = 1 will be 0.84134
Probability for Z = - 1 will be 0.15866
Therefore , probability that products are within specs = 0.84134 – 0.15866 = 0.68268
Therefore probability that products are outside specs = 1 – 0.68268 = 0.31732 ( 31.732 %)
THE FRACTION OF PRODUCTS THAT IS OUT OF SPECS IS ABOUT 30%
ANSWER : ALL OF THE ABOVE
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