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
and 
, aprox -19.1168 and -1.8831