Answer:
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
Step-by-step 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
