a)
b))
mean E(x)=μ=np=0.4 |
expected number =0.4
c)
P(batch will be accepted) =P(X<=1)= | ∑x=0a (nCx)px(1−p)(n-x) = | 0.943 |
d)
standard deviation σ=√(np(1-p))=0.616 |
the second image is part of the question, please answer as much of the question you...
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