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    				| 1. | 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 fewer than 2 defects. | 
| Answer» Given mean np = 1.2 and n = 6 p = 1.2/6 = 0.2, q = 1 – 0.2 = 0.8 Let X be a binomial variable denoting the number of defects, (i.e,) X ~ B (6, 0.2) p.m.f is given by P (X = x) = 6Cx(0.2)x(0.8)6 - x We want P(X < 2) = P(X = 0) + P (X = 1) = 6C0(0.2)0(0.8)6 + 6C1(0.2)1 (0.8)5 = (0.8)6 + 6 (0.2) (0.8)5 = 0.262144 + 0.393216 = 0.65536 Thus if lengths of 6 metres are to be inspected, the probability of less than 2 defects is 0.65536. | |