When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test56batteries and determine whether each is within specifications. The entire shipment is accepted if at most2batteries do not meet specifications. A shipment contains5000​batteries, and11​%of them do not meet specifications. What is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?The probability that this whole shipment will be accepted is?​(Round to four decimal places as​ needed.)The company will acceptwhat %of the shipments and will rejectwhat ​%of the​ shipments, so

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Asked Oct 17, 2019
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When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test
56
batteries and determine whether each is within specifications. The entire shipment is accepted if at most
2
batteries do not meet specifications. A shipment contains
5000
​batteries, and
11​%
of them do not meet specifications. What is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?
The probability that this whole shipment will be accepted is?
​(Round to four decimal places as​ needed.)
The company will accept
what %
of the shipments and will reject
what ​%
of the​ shipments, so
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Expert Answer

Step 1

Binomial distribution:

Since there are only two outcomes (shipment accepted and shipment rejected), the given scenario follows binomial distribution. The pmf of binomial distribution is as follows:

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P(X x) "C,pq* where, n is the number of trials, p is the probability of success, and q is the probability of failure

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Step 2

Computation of percentage of shipments that the company will accept:

Here, n = 56

Probability of success, p = 0.11

Probability of failure, q = 0.89 (= 1−0.11)

The entire shipment is accepted if at most 2 batteries do not meet specification. That is we have to find P(x ≤ 2).

The percentage of shipments that the company will accept is obtained as 4.61%, from the calculations given below:

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P(x2) P(x 0) + P(x=1)+ P(x=2) C,(0.11) (0.89)*C(0.11) (0.89)*C(0.11 (0.89) *? 56-1 56-2 = 0.0015 0.0101 0.0345 =0.0461 =4.61%

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Step 3

Computation of percentage of shipments that the company will reject:

The percentage of shipments that the compan...

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The shipments that the P 1-0.0461 company will reject 0.9539 = 95.39%

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