Respuesta :
Using the binomial distribution, it is found that there is a:
a) 0.3526 = 35.26% probability of rejecting a lot that is 3% defective.
b) 0.4295 = 42.95% probability of accepting a lot that is 4% defective.
For each device, there are only two possible outcomes, either it is defective, or it is not. The probability of a device being defective is independent of any other device, hence the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem, the sample has 100 units, hence [tex]n = 100[/tex].
Item a:
3% of the pieces are defective, hence [tex]p = 0.03[/tex].
The probability is:
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which:
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
Hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.03)^{0}.(0.97)^{100} = 0.0476[/tex]
[tex]P(X = 1) = C_{100,1}.(0.03)^{1}.(0.97)^{99} = 0.1471[/tex]
[tex]P(X = 2) = C_{100,2}.(0.03)^{2}.(0.97)^{98} = 0.2252[/tex]
[tex]P(X = 3) = C_{100,3}.(0.03)^{3}.(0.97)^{97} = 0.2275[/tex]
Then:
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0476 + 0.1471 + 0.2252 + 0.2275 = 0.6474[/tex]
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.6474 = 0.3526[/tex]
0.3526 = 35.26% probability of rejecting a lot that is 3% defective.
Item b:
4% of the pieces are defective, hence [tex]p = 0.04[/tex].
Lot is accepted if less than 4 units are defective, hence:
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = 0) = C_{100,0}.(0.04)^{0}.(0.96)^{100} = 0.0169[/tex]
[tex]P(X = 1) = C_{100,1}.(0.04)^{1}.(0.96)^{99} = 0.0703[/tex]
[tex]P(X = 2) = C_{100,2}.(0.04)^{2}.(0.96)^{98} = 0.1450[/tex]
[tex]P(X = 3) = C_{100,3}.(0.04)^{3}.(0.96)^{97} = 0.1973[/tex]
Then:
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0169 + 0.0703 + 0.1450 + 0.1973 = 0.4295[/tex]
0.4295 = 42.95% probability of accepting a lot that is 4% defective.
A similar problem is given at https://brainly.com/question/24863377
