Question

A manufacturer produces lots of a canned food product. Let p denote the proportion of the lots that do not meet the product quality specifications. An

n = 29, c = 0

acceptance sampling plan will be used.

(a)

Compute points on the operating characteristic curve when

p = 0.01, 0.03, 0.10, and 0.20.

(Round your answers to four decimal places.)

c	p = 0.01	p = 0.03	p = 0.10	p = 0.20
0

(c)

What is the probability that the acceptance sampling plan will reject a lot containing 0.10 defective? (Round your answer to four decimal places.)

Answer 1

SOLUTION:

From given data,

A manufacturer produces lots of a canned food product. Let p denote the proportion of the lots that do not meet the product quality specifications.

n = 29, c = 0

(a) Compute points on the operating characteristic curve when

p = 0.01, 0.03, 0.10, and 0.20.

(i) n = 29 , p = 0.01 and c = 0

P(X < 0 ) = P(X = 0)

P(X < 0 ) = ²⁹C₀  (0.01)⁰  (1-0.01)^35-0

P(X < 0 ) = 1 * 1 * (0.99)³⁵

P(X < 0 ) = 0.7034

(ii) n = 29 , p = 0.03 and c = 0

P(X < 0 ) = P(X = 0)

P(X < 0 ) = ²⁹C₀  (0.03)⁰ (1-0.03)^35-0

P(X < 0 ) = 1 * 1 * (0.97)³⁵

P(X < 0 ) =0.3443

(iii) n = 29 , p = 0.10 and c = 0

P(X < 0 ) = P(X = 0)

P(X < 0 ) = ²⁹C₀  (0.10)⁰  (1-0.10)^35-0

P(X < 0 ) = 1 * 1 * (0.9)³⁵

P(X < 0 ) = 0.0250

(iv) n = 29 , p = 0.20 and c = 0

P(X < 0 ) = P(X = 0)

P(X < 0 ) = ²⁹C₀ (0.20)⁰  (1-0.20)^35-0

P(X < 0 ) = 1 * 1 * (0.8)³⁵

P(X < 0 ) = 0.0004

C	p = 0.01	p = 0.03	p = 0.10	p = 0.20
O	0.7034	0.3443	0.0250	0.0004

(c) What is the probability that the acceptance sampling plan will reject a lot containing 0.10 defective? (Round your answer to four decimal places.)

1 -P(X < 0 )  = 1 - P(X = 0)

= 1 - 0.7034

= 0.2966

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