What is the volume whose 16 whole 2/3 %is 0.75 litre
1 answer:
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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
<h3>P(x)=∑ᵇₐ=₀ (n!/a!(n-a)!)p^a (1-p)^(n-a)</h3>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
=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!