Question

3.8.10) Inclusions are defects in poured metal caused by contamitants. The number of (large) inclusions in cast iron follows a Poisson distribution with a mean of 2 per cubic millimeter. Deter mine the following:

(a) Probability of at least one inclusion in a cubic millimeter.
(b) Probability of at least 3 inclusions in 4.0 cubic millimeters.
(c) Volume of material to inspect such that the probability of at least one inclusion is 0.98
(d) Determine the mean inclusion per cubic millimeter such that the probability of at least one inclusion is 0.98.

Answer 1

a) P(X>=1) = 0,7293 or P(X>=1) = 72,93 %

b) P(X>=3) = 0,5686 or P(X>=3) = 56,86 %

c) y = 0,8494 mm³

d) λ = 1,69897 inc per mm³

Explanation:

a) P(X>=1) = 1 - P(X=0)

P(X = 0) = λ⁰ * e∧⁻λ / 0! where λ = 2 in/mm³

P(X = 0) = e∧-2 ⇒ P(X = 0) = 0,2706

Then

P(X>=1) = 1 - 0,2706

P(X>=1) = 0,7293 or P(X>=1) = 72,93 %

b) 3 inc. in 4 mm³ = 2*4/3 λ = 2,67 inc. per mm³

P(X>=3) = 1 - P(X=1 ) - P(X=2)

P(X=1) = λ¹ * e∧-λ / 1! ⇒ P(X=1) = 2,67 *e∧- 2,67/1

P(X=1) = 0,1846

P(X=2) = λ² * e∧-λ / 2! ⇒ P(X=2) = (2,67)² * e ∧ - 2,67/2

P(X=2) = 7,1289* 0,06925/2 ⇒ P(X=2) = 0,2468

Then

P(X>=3) = 1 - 0,1846 - 0,2468

P(X>=3) = 0,5686 or P(X>=3) = 56,86 %

c) P(X>=1) = 0,98

and λ = 2*y where y is the quantity of material

P(X=0) = 1 - P(X>=1)

P(X=0) = 1 - 0,98 P(X=0) = 0,02

0,02 = (2*y)⁰ * e ∧ -2*y /0!

0,02 = 1 * e ∧ -2*y

Taking log on both sides of the equation

log (0,02) = -2*y - 1,69897 = - 2*y

y = 0,8494 mm³

d) P(X >=1 ) = 1 - P( X = 0)

0,98 = 1 - P( X = 0)

P( X = 0) = 0,02

0,02 = λ⁰ * e ∧ - λ / 0!

0,02 = e ∧ -λ

Taking log on both sides of the equation we get:

log ( 0,02 ) = - λ

- 1,69897 = - λ

λ = 1,69897 inc. per mm³