Respuesta :
Using the binomial distribution, it is found that the smallest number of flaked stone objects that need to be found to be at least 90% sure that at least one is a finished arrow point is 22.
For each object, there are only two possible outcomes. Either it is a finished arrow point or not. The probability of an object being a fixed arrow point is independent of any other object, which means that the binomial distribution is used to solve this question.
Binomial 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, 10% of the objects were finished arrow points, thus [tex]p = 0.1[/tex].
We want to be at least 90% sure that at least one is a finished arrow point, that is:
[tex]P(X \geq 1) \geq 0.9[/tex]
Considering [tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
[tex]1 - P(X = 0) \geq 0.9[/tex]
[tex]P(X = 0) \leq 0.1[/tex]
We have that:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{n,0}.(0.1)^{0}.(0.9)^{n} = (0.9)^n[/tex]
Then, we can solve for n:
[tex]P(X = 0) \leq 0.1[/tex]
[tex](0.9)^n \leq 0.1[/tex]
[tex]\log{(0.9)^n} \leq \log{0.1}[/tex]
[tex]n\log{0.9} \geq \log{0.1}[/tex]
[tex]n \geq \frac{\log{0.1}}{\log{0.9}}[/tex]
[tex]n \geq 21.9[/tex]
Rounding up, the smallest number of flaked stone objects that need to be found to be at least 90% sure that at least one is a finished arrow point is 22.
A similar problem is given at https://brainly.com/question/22648840
