Respuesta :
Answer:
a) [tex] P(X=0)[/tex]
Who represent 0 defective and using the probability mass function we got:
[tex] P(X=0) = (4C0) (0.0367)^0 (1-0.0367)^{4-0} = 0.8612[/tex]
b) [tex] P(X=0) = (100C0) (0.0367)^0 (1-0.0367)^{100-0} = 0.02378[/tex]
For this case we can conclude that the outlet plan would deal the defectives returned by consumers
Step-by-step explanation:
Let X the random variable of interest "number of defectives", on this case we now that:
[tex]X \sim Binom(n, p)[/tex]
the probability of being defective for this case is given by:
[tex] p = \frac{220}{6000}= 0.0367[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
For this case n = 4 and we want to find this probability:
[tex] P(X=0)[/tex]
Who represent 0 defective and using the probability mass function we got:
[tex] P(X=0) = (4C0) (0.0367)^0 (1-0.0367)^{4-0} = 0.8612[/tex]
Part b
For this case n = 100 and we want to find this probability:
[tex] P(X=0)[/tex]
Who represent 0 defective and using the probability mass function we got:
[tex] P(X=0) = (100C0) (0.0367)^0 (1-0.0367)^{100-0} = 0.02378[/tex]
For this case we can conclude that the outlet plan would deal the defectives returned by consumers
