An inspector selects five parts (without replacement) from each batch of parts p

Question

An inspector selects five parts (without replacement) from each batch of parts produced by a production line; the batch fails inspection if at least one of these five parts is defective (a)Compute the probability that a batch does not fail when it contains 30 parts, of which six are defective (b)Compute the probability that a batch fails inspection when it contains 90 parts, of which ten are defective (c) If one batch like the one described in Part (b) is inspected each day, what is average number of days until the first failed inspection? (d)Consider the batch from Part (b). Compute the mean and variance of the number of non-defective parts among those parts selected by the inspector Consider the batch from Part (b). Suppose the inspector will select t parts for inspection (not necessarily five; the batch still fails if at least one of these t parts is defective). Find the smallest value of t that will cause this batch to fail inspection with probability 0.70 or higher (e)

Explanation / Answer

Ans:

Binomial distribution

a)n=30

p=6/30=1/5=0.2

P(k>=1)=1-P(k=0)=1-5C0*0.20*0.85

=1-0.85

=1-0.3277

=0.6723

b)p=10/90=1/9=0.11

P(k>=1)=1-5C0*0.110*0.895

=1-0.895

=1-0.5584

=0.4416

c)Geometric distribution with p=0.4416

Mean=1/0.4416=2.26

i.e it will take on average 2.26 days to until the first inspection failed.

d)mean=np=5*0.89=4.45

variance=np(1-p)=4.45*0.11=0.4895

e)

Hence,t>=11,which will cause this batch to fail inspection ith probability 0.7 or higher.

t 1-BINOMDIST(0,t,0.11,FALSE) 5 0.4416 6 0.5030 7 0.5577 8 0.6063 9 0.6496 10 0.6882 11 0.7225

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