IGNOU MSTE 1 2018 - A leather bag manufacturing company supplies bags in lots of size 150

Question

A leather bag manufacturing company supplies bags in lots of size 150 to a buyer. A single sampling plan with n = 10 and c = 1 is being used for the lot inspection. The company and the buyer decide that AQL = 0.08 and LTDD = 0.16. If there are 15 defective bags in each lit, compute the

i) Probability of accepting the lot,

ii) Producer’s risk and consumer’s risk,

iii) AOQ, if the rejected lots are screened and all defective bags are replaced by non- defectives, and

iv) Average total inspection

16 Feb 2018

Answer :

Word Count : 632

Let's solve this step by step manually, using the single sampling plan formulas. Here's what we have: Given: * Lot size: $N = 150$ * Sample size: $n = 10$ * Acceptance number: $c = 1$ * AQL = 0.08 * LTPD = 0.16 * Number of defectives in lot: $15$ First, calculate the fraction defective in the lot: $$ p = \frac{\text{Number of defectives}}{\text{Lot size}} = \frac{15}{150} = 0.1 $$ --- ### Step (i) Probability of accepting the lot For a single sampling plan, probability of accepting the lot ($P_a$) is the probability that at most $c$ defectives are in the sample. This follows the hypergeometric distribution: $$ P_a = \sum_{x=0}^{c} \frac{\binom{D}{x}\binom{N-D}{n-x}}{\binom{N}{n}} $$ Where: * $D = 15$ defectives, $N = 150$, $n = 10$, $c = 1$ We need: $$ P_a = P(X = 0) + P(X = 1) $$ Step 1: $P(X = 0)$ $$ P(X = 0) = \frac{\binom{15}{0} \binom{135}{10}}{\binom{150}{10}} $$ $\binom{15}{0} = 1$ Approximate using binomial approximation, since $n/N$ is small ($10/150 = 0.0667$): $$ P(X=0) \approx (1 - p)^n = (1 - 0.1)^{10} = 0.9^{10} $$ $$ 0.9^{10} \approx 0.3487 $$ Step 2: $P(X = 1)$ $$ P(X = 1) \approx \binom{10}{1} (0.1)^1 (0.9)^9 = 10 \cdot 0.1 \cdot 0.3874 \approx 0.3874 $$

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