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 7000 batteries, and 1% 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?
Accepted Solution
A:
Answer:98.1% chance of being acceptedStep-by-step explanation:Given:sample size,n=56acceptance condition= at most 2 batteries do not meet specificationsshipment size=7000battery percentage in shipment that do not meet specification= 1%Applying binomial distributionP(x)=∑ᵇₐ=₀ (n!/a!(n-a)!)p^a (1-p)^(n-a)In this formula, a is the acceptable number of defectives; n is the sample size; p is the fraction of defectives in the population. Now putting the valuea= 2 n=56p=0.01[tex]\frac{56!}{0!\left(56-0\right)!}\left(0.01\right)^0\:\left(1-0.01\right)^{\left(56-0\right)} + \frac{56!}{1!\left(56-1\right)!}\left(0.01\right)^1\:\left(1-0.01\right)^{\left(56-1\right)} +[/tex][tex]\:\frac{56!}{2!\left(56-2\right)!}\left(0.01\right)^2\:\left(1-0.01\right)^{\left(56-2\right)}[/tex]=0.56960+0.32219+0.08949After summation, we get 0.981 i.e. a 98.1% chance of being accepted. As this is such a high chance, we can expect many of the shipments like this to be accepted!