Question

A factory bought a new machine. It is not configured well yet, so the probability of defects is 0.05. Estimate the probability that there are at most 2 spoiled details in the random sample of 20.

Answer 1

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 2

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