Question

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?

Answer 1

Answer:

98.1% chance of being accepted

Step-by-step explanation:

Given:

sample size,n=56

acceptance condition= at most 2 batteries do not meet specifications

shipment size=7000

battery percentage in shipment that do not meet specification= 1%

Applying binomial distribution

P(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 value

a= 2

n=56

p=0.01

$\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)} +$$\:\frac{56!}{2!\left(56-2\right)!}\left(0.01\right)^2\:\left(1-0.01\right)^{\left(56-2\right)}$

=0.56960+0.32219+0.08949

After 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!