Question #103651
When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 48 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 3 batteries do not meet specifications. A shipment contains 5000 ​batteries, and 2​% 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
nothing.
​(Round to four decimal places as​ needed.)
1
Expert's answer
2020-02-25T08:27:17-0500

Let X=X= the number of defective batteries. The probability distribution of the random variable X,X, hypergeometric distribution, is given by


h(x;n,M,N)=(Mx)(NMnx)(Nn)h(x;n, M, N)={\binom{M}{x}\binom{N-M}{n-x} \over \binom{N}{n}}

Given that N=5000,M=50000.02=100,n=48.N=5000, M=5000\cdot0.02=100,n=48.

What is the probability that at most 3 batteries do not meet specifications?


P(X3)=P(X=0)+P(X=1)+P(X\leq3)=P(X=0)+P(X=1)++P(X=2)+P(X=3)=+P(X=2)+P(X=3)==(1000)(5000100480)(500048)+(1001)(5000100481)(500048)+={\binom{100}{0}\binom{5000-100}{48-0} \over \binom{5000}{48}} +{\binom{100}{1}\binom{5000-100}{48-1} \over \binom{5000}{48}}++(1002)(5000100482)(500048)+(1003)(5000100483)(500048)+{\binom{100}{2}\binom{5000-100}{48-2} \over \binom{5000}{48}}+{\binom{100}{3}\binom{5000-100}{48-3} \over \binom{5000}{48}}\approx0.377432+0.373310+0.178926+0.055379\approx0.377432+0.373310+0.178926+0.055379\approx0.9850\approx0.9850

The probability that this whole shipment will be accepted is  0.9850.0.9850.



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