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A marketing study contained the following questions(i) Do you own.pdf
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### A manufacturer buys threaded bolts in large quantities from two supp.pdf

1. A manufacturer buys threaded bolts in large quantities from two suppliers S1 and S2. The bolts are kept in a bin from where they are drawn. 40% of the bolts are from S1 and 60% from S2. 1.5% of S1 bolts are bad and .5% of bolts from S2 are bad. If bolts from S1 are inspected until one bolt that is NOT bad is found, what is the expected number of bolts that must be inspected? My thoughts is this is the mean for a binomial and it would be n*p? Not sure of how to actually work this one out. Solution This is a geometric distribution. You are sampling until a bolt is not bad. If you would like to derive the distribution, let p be the probability that the first bolt is good. Trivially, then, p(1) = p. To determine p(n), if n is the first time the bolt is not bad, the first n-1 bolts had to be bad. The probability that the first n-1 bolts are bad = (1-p) ^(n-1). The probability that the n'th bolt is good is p. Thus, p(n) = (1-p) ^(n-1) * p. Note that this formula also works for n = 1. The mean of a geometric distribution with probability p is 1/p. 1/(1-.015) = 1.01522842639594 . We might want to express this as 1.0152. See http://en.wikipedia.org/wiki/Geometric_distribution for an explanation. (We are using the first definition.)
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