A manufacturer assures his customers that the probability of having defective item is as 0.005. A sample of 1000 items was inspected. Determine the probabilities of having the given possible outcomes

i. Simply one is defective

ii. Mostly 2 defective

iii. More than 3 defective

P(x) =  e^-λ λ^x/x!

(? = np = 1000 × 0.005) = 5

i. P(only one is defective) = P(1) = P(x = 1)

= (2.718^-5 * 5¹)/1!     Note that 2.718^-5=  1/(2.718)⁵

= 5/(2.718)⁵

= 5/(148.33)

ii. P(at most 2 defective) = P(x ≤ 2)

= P(0) +P(1)+P(2)

P(x=0) = e^-55⁰/0!

= 2.718^-5

= 1/(2.718)⁺⁵

= 0.00674

P(1) = 0.0337

P(2) = (2.718^-5 * 5²)/2!

= 0.08427

P(x<2) = 0.00674 + 0.0337 + 0.08427

= 0.012471

iii. P(more than 3 defective) = P(x > 3)

= 1 -  [P(0)+P(1)+P(2)+P(3)]

Posted Date: 2/20/2013 6:25:51 AM | Location : United States

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