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Question
Defects in yarn manufactured by a local mill can be approximated by a distribution with a mean of 1.2 defects for every 6 metres of length. If lengths of 6 metres are to be inspected, find the probability of less than 2 defects
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Solution
Given n = 6
Mean np = 1.2
⇒ 6p = 1.2
p = `1.2/6 "p" = 0.2`or p = `1/5`
q = 1 – p = `1 - 1/5`
∴ q = `4/5`
The binomial distribution
p(x) = 6Cx (0.2)x(0.8)6-x
p(x < 2) = p(x = 0) + p(x = 1)
= `6"c"_0 (1/5)^0 (4/5)^(6 - 0) + 6"c"_1 (1/5)^1 (4/5)^(6 - 1)`
= `(1)(1)(4/5)^6 + 6(1/5)(4/5)5`
= `(4)^6/(5)^6 + (6 xx (4)^5)/(5)^6`
= `4096/15625 + ((6 xx 1024)/(15625))`
= `(4096 + 6144)/15625`
= `10240/15625`
= 0.65536
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