# Industrial Engineering Questions and Answers – Inspection Sampling Plans – Set 2

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

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

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

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

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

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

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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