Respuesta :
Answer:
There is a 17.09% proability that on any given day the production process will be stopped
Step-by-step explanation:
The production will be stopped if there are two or more defective units, so we have to find [tex]P(X \geq 2)[/tex].
This is [tex]P(X \geq 6)[/tex]. We know that either we have less than 2 defecive units, or we have at least 2 defective units. The sum of the probabilities is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
So, we have to find P(X < 2), that is:
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
Since either an unit is defective or it is not, that is, there are ony two possible outcomes, P(X = 0) and P(X = 1) are binomial probabilities.
Binomial probability
Th binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And [tex]\pi[/tex] is the probability of X happening.
So:
The probability of a defective unit is 0.05, so [tex]\pi = 0.05[/tex]
The size of the sample is 15, so [tex]n = 15[/tex]
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 0) = C_{15,0}.(0.05)^{0}.(0.95)^{15} = 0.4633[/tex]
[tex]P(X = 1) = C_{15,1}.(0.05)^{1}.(0.95)^{14} = 0.3658[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.4633 + 0.3658 = [/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8291 = 0.1709[/tex]
There is a 17.09% proability that on any given day the production process will be stopped
