Respuesta :
Answer:
The probability of getting two or more defectives in a sample of 200 items
0.3233
Step-by-step explanation:
step1:-
The probability of getting 1 percentage defective items is
P(D) = [tex]\frac{1}{100} =0.01[/tex]
Given total number of items is n=200
use the Poisson approximation to the binomial distribution that is
λ=np=200(0.01)=2
by using Poisson distribution P(X=r)=[tex]e^-λ (λ^ r)[/tex]
Step2:-
The probability of getting two or more defectives items
[tex]P(x\geq 2)=1-{P(x\leq 2)[/tex]
[tex]P(x\geq 2)=1-{P(x=0)+P(x=1)+P(x=2)][/tex]
using Poisson distribution
[tex]P(x\geq 2)=1-[{e^{-2} \frac{2^{0} }{0!} } +e^{-2} \frac{2^{1} }{1!} +e^{-2} \frac{2^{2} }{2!} ][/tex]
on simplification, we get
[tex]P(x\geq 2)=1-({e^{-2})(1+2+2)[/tex]
[tex]P(x\geq 2)=1-({e^{-2})5[/tex]
by using calculator e^-2 value is 0.13533
[tex]P(x\geq 2)=1-({0.13533)X5[/tex]
after simplification
[tex]P(x\geq 2)=1-0.676676[/tex]
[tex]P(x\geq 2)= 0.3233[/tex]
Final answer:-
The probability of getting two or more defectives in a sample of 200 items
0.3233
