This set of Industrial Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Inspection Sampling Plans – Set 2”.
1. Which sampling plan is more acceptable to the consumer?
a) Double sampling plan
b) Single sampling plan
c) Sequential sampling plan
d) Multiple sampling plans
View Answer
Explanation: Single sampling plan is more acceptable to the consumer because it involves more rejection of lots which results in 100% inspection of goods in a lot.
2. Find the probability of containing one defective in the sample of 15. Take lot size as 50 articles of which 2 % are defectives.
a) 0.1
b) 0.2
c) 0.5
d) 0.3
View Answer
Explanation:
Given,
Sample size, n = 15
Lot size, N = 50
Total No of defectives = 0.02 × 50 = 1
No of non – defectives items = 50 – 1 = 49
Probability of containing one defective article in a sample of 15 is \(\frac {49C14 \times 1C1}{50C15} = \frac {15}{50}\) = 0.3
3. A double sampling plan is more acceptable to the producer.
a) True
b) False
View Answer
Explanation: A double sampling plan is more acceptable to the producer because it involves inspection of two samples which mostly results in fewer lot rejections. Hence, 100% inspection of lots can be minimised to the maximum extent possible.
4. Find the probability of acceptance of a lot of 50 articles having 2% as defectives using a single sampling plan. Take lot size as 15 and acceptance number as 1.
a) 0.1
b) 0.01
c) 1
d) 0.001
View Answer
Explanation:
Given
Lot size, N = 50
Sample size, n = 15
Acceptance Number, C = 1
Total No of defectives = 0.02 × 50 = 1
No of non – defectives items = 50 – 1 = 49
Probability of containing zero defective articles in a sample of 15 is Po = \(\frac {49C15 \times 1C0}{50C15} = \frac {35}{50}\) = 0.7
Probability of containing one defective article in a sample of 15 is P1 = \(\frac {49C14 \times 1C1}{50C15} = \frac {15}{50}\) = 0.3
Therefore, the total probability of acceptance = Po + P1 = 0.7 + 0.3 = 1.
5. Calculate the probability of acceptance (Pa) for a lot of size 150 of which 4% are defectives. Take n1 = 5, n2 = 5, c1 = 0, c2 = 2, r1 = 2, r2 = 3.
a) 0.981
b) 0.859
c) 0.189
d) 0.891
View Answer
Explanation:
Given,
N = 150
No of defectives items = \(\frac {4}{100}\) × 150 = 6
No of non-defectives items = 144
The lot will be accepted when
- 1st sample contains 0 defectives (or)
- 1st sample contains one defective item and 2nd sample contains no defective item (or)
- 1st sample contains one defective item and 2nd sample contains one defective item
Probability of 1st sample contains 0 defectives is \(\frac {144C5 \times 6C0}{150C5}\) = 0.813
Probability of 1st sample contains one defective item and 2nd sample contains no defective item is \(\frac {144C4 \times 6C1}{150C5} \times \frac {140C5 \times 5C0}{145C5}\) = 0.174 × 0.839 = 0.146
Probability of 1st sample contains one defective item and 2nd sample contains no defective item is \(\frac {144C4 \times 6C1}{150C5} \times \frac {140C4 \times 5C1}{145C5}\) = 0.174 × 0.131 = 0.022
Total probability of acceptance (Pa) = 0.813 + 0.146 + 0.022 = 0.981.
6. Calculate AOQ for a lot of size 10000 from which a sample of size 100 is drawn for inspection using a single sampling plan with acceptance number, C = 2. Take 1% of lot as defectives and Pa = 0.92
a) 0.092
b) 0.0092
c) 0.92
d) 0.0902
View Answer
Explanation:
Given,
N = 10000
N = 100
C = 2
Pa = 0.92
AOQ = P × Pa = 0.01 × 0.92 = 0.0092.
Sanfoundry Global Education & Learning Series – Industrial Engineering.
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