Respuesta :
Answer:
0.1902 = 19.02% probability of returning a shipment
Step-by-step explanation:
For each alternator, there are only two possible outcomes. Either it is defective, or it is not. The probability of an alternator being defective is independent of any other alternator, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The 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}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations 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 p is the probability of X happening.
1.5% of the alternators will not perform up to specifications.
This means that [tex]p = 0.015[/tex]
Shipment of 100 alternators
This means that [tex]n = 100[/tex]
What is the probability of returning a shipment?
Probability of more than 2 defective, which is:
[tex]P(X > 2) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.015)^{0}.(0.985)^{100} = 0.2206[tex]
[tex]P(X = 1) = C_{100,1}.(0.015)^{1}.(0.985)^{99} = 0.3360[/tex]
[tex]P(X = 2) = C_{100,2}.(0.015)^{2}.(0.985)^{98} = 0.2532[/tex]
Then
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2206 + 0.3360 + 0.2532 = 0.8098[/tex]
Then
[tex]P(X > 2) = 1 - P(X \leq 2) = 1 - 0.8098 = 0.1902[/tex]
0.1902 = 19.02% probability of returning a shipment
