Question

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

Answer and Step-by-step explanation:

Solution: given:

(a)

Mean = λ = µ = 2permm3

By using formula:

P(x = k) =  λk e- λ  / k!

P ( not A) = 1 – p(A)

Passion probability at k = 0

P( x = 0) = 20 e-2/ 0!

               = 0.1353

Probability of at least one inclusion:

P(x≥1) = 1- p(x=0) = 1 – 0.1353

            = 0.864

(b) Probability of at least 3 inclusions in 4.0 cubic millimeters.

Mean = λ = µ = 2 per mm3

Inclusion in 4 cubic mm = 4 x 2 per 4 mm3 = 8 per 4 mm3

P(not A) = 1 – p(A)

Passion probability at k = 0,1,2

P(x=0) = 80 e-8 / 0!

        = 0.00034

P(x = 1) = 81 e-8 / 1!

            = 0.00272

P(x = 2) = 82 e-8 / 2!

              = 0.01088

Probability at least 3 inclusion:

P(x≥3) = 1 – p(x<3) = 1 – p(x = 0) – p(x =1) – p(x = 2)

            = 1 – 0.00034 – 0.00272 – 0.01088

          = 0.98606

(c) . Volume of material to inspect such that the probability of at least one inclusion is 0.98

The probability of x ≥ 1 has to be equal to 0.98

P(x ≥ 1) = 0.98

P(x = 0) = p(x < 1)

            = 1 – p(x≥ 1) = 1 – 0.98

            = 0.02

Poission probability at k = 0

P(x = 0) = (2 n )0 e-2n / 0! = e-2n

e-2n = 0.02

taking ln on both sides :

-2 n = ln e-2n = ln 0.02

Divided by -2,

n = ln 0.02 / -2

    = 1.955

Volume is 1.955 mm3.

(d) Determine the mean inclusion per cubic millimeter such that the probability of at least one inclusion is 0.98.

The probability of x ≥ 1has to be at least 0.98

P(x ≥ 1 )≥ 0.98

P(x = 0 ) = p(x < 1 ) = 1 – p(x≥1) =  1- 0.98

           = 0.02

Poisson probability at k = 0

P(x = 0) = λ0 e-λ / 0! = e-λ

e-λ ≤ 0.02

taking ln on both sides:

ln e-λ ≤ ln 0.02

λ = ln 0.02 / -1

    = 3.912