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QUESTION 14 The number of flaws in a given area of aluminum folil is 3 per m2 Find the probability that there are no flaws in

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Answer #1

Since the average number of flaws that occurs in a specified region (area) is known and the probability that a flaw will occur is proportional to the size of the region (area), the number of flaws in a given area can be modeled as Poisson distribution.

Let X be the number of flaws in 1 m2

Then X ~ Poisson(\lambda = 3)

14.

Probability of no flaw = P(X = 0) = exp(-3) * 30 / 0!

= exp(-3)

= 0.05

15.

Probability of more than 3 flaws = P(X > 3)

= 1 - P(X \le 3)

= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= 1 - [exp(-3) * 30 / 0! + exp(-3) * 31 / 1! + exp(-3) * 32 / 2! + exp(-3) * 33 / 3! ]

= 1 - (0.04978707 + 0.14936121 + 0.22404181 + 0.22404181)

= 0.353

16.

Rate \lambda for 1 square meter = 3

Rate \lambda for 2 square meters = 3 * 2 = 6

17.

For 2 square meters, \lambda = 6

Probability that there are at least 8 flaws = P(X \ge 8)

= 1 - P(X < 8)

= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)]

= 1 - [exp(-6) * 60 / 0! + exp(-6) * 61 / 1! + exp(-6) * 62 / 2! + exp(-6) * 63 / 3! + exp(-6) * 64 / 4! + exp(-6) * 65 / 5! + exp(-6) * 66 / 6! + exp(-6) * 67 / 7!]

= 1 - [0.002478752 + 0.014872513 + 0.044617539 + 0.089235078 + 0.133852618 + 0.160623141 + 0.160623141 + 0.137676978]

= 0.256

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