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 P

_{o}= \(\frac {49C15 \times 1C0}{50C15} = \frac {35}{50}\) = 0.7

Probability of containing one defective article in a sample of 15 is P

_{1}= \(\frac {49C14 \times 1C1}{50C15} = \frac {15}{50}\) = 0.3

Therefore, the total probability of acceptance = P

_{o}+ P

_{1}= 0.7 + 0.3 = 1.

5. Calculate the probability of acceptance (P_{a}) for a lot of size 150 of which 4% are defectives. Take n_{1} = 5, n_{2} = 5, c_{1} = 0, c_{2} = 2, r_{1} = 2, r_{2} = 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

- 1
^{st}sample contains 0 defectives (or) - 1
^{st}sample contains one defective item and 2^{nd}sample contains no defective item (or) - 1
^{st}sample contains one defective item and 2^{nd}sample contains one defective item

Probability of 1^{st} sample contains 0 defectives is \(\frac {144C5 \times 6C0}{150C5}\) = 0.813

Probability of 1^{st} sample contains one defective item and 2^{nd} 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 1^{st} sample contains one defective item and 2^{nd} 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 (P_{a}) = 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 P_{a} = 0.92

a) 0.092

b) 0.0092

c) 0.92

d) 0.0902

View Answer

Explanation:

Given,

N = 10000

N = 100

C = 2

P

_{a}= 0.92

AOQ = P × P

_{a}= 0.01 × 0.92 = 0.0092.

**Sanfoundry Global Education & Learning Series – Industrial Engineering**.

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