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Question
A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain at least 2 rejects?
Solution
In a binomial distribution
n = 10
p = `12/100 = 3/25`
q = – p = `1 - 3/25` q = = `22/25`
PX = x) = ncxpxqn-x
P(at least 2 rejects) = P(X ≥ 2)
= `1 - "P"("X" < 2)`
= `1 - ["P"("X" = 0) + "P"("X" = 1)]`
= `1 - [(22)^10/(25)^10 + (30 xx (22)^9)/(25)^10]`
= `1 - {(22)^9/(25)^10 [22 + 30]}`
= `1 - {(52 xx (22)^9)/(25)^10}`
= `1 - {(52 xx 1.207 xx 10^12)/(9.528 xx 10^13)}`
= `1 - [(52 xx 1.207)/(9.528 xx 10)]`
= ` -[62.764/95.28]`
= `1 - 0.6587`
= 0.3413
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