Respuesta :
Answer:
Pr(at most 2 are defective) = 0.38
Step-by-step explanation:
probability of defective = 0.05
probability of not defective =1- 0.05 = 0.95
Pr(at most 2 are defective) = pr (0 defective) + pr (1 defective) + pr (2defective)
pr (0 defective) = (0.05^0) * (0.95^20) = 0.36
pr (1 defective) = (0.05^1) * (0.95^19) = 0.019
pr (2 defective) = (0.05^2) * (0.95^18) = 0.0025 * 0.397 = 0.00099
Pr(at most 2 are defective) = 0.36 + 0.019 + 0.00099
Pr(at most 2 are defective) = 0.37999 = 0.38
Answer:
P(x<2)=0.55
Step-by-step explanation:
Using poison binomial distribution
For pois(landa)
N=20
P=0.05
X=2
Now we will use (λ) which is the parameter of poisson distribution
(λ)=np=0.05*20=1
Now since we want since probability of getting at most 2 spoiled
P(x>2)=1- P(x<2)=1(p(x=2)+p(x=1)+p(x=0)
e^-(λ) λ^k/
k!
So therefore λ=1
(e-1 1^2/2!) + (e-1 1^1/1!) + (e-1 1^0/0!)
Equals 0.18+0.37+0=0.55