IDNLearn.com is designed to help you find the answers you need quickly and easily. Ask anything and get well-informed, reliable answers from our knowledgeable community members.
Sagot :
Using the binomial distribution, it is found that there is a 0.7447 = 74.47% probability that fewer than 7 of them show up.
What is the binomial distribution formula?
The formula is:
[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 values of the parameters are given as follows:
p = 0.7, n = 8.
The probability that fewer than 7 of them show up is given by:
[tec]P(X < 7) = 1 - P(X \geq 7)[/tex]
In which:
[tex]P(X \geq 7) = P(X = 7) + P(X = 8)[/tex]
Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{8,7}.(0.7)^{7}.(0.3)^{1} = 0.1977[/tex]
[tex]P(X = 8) = C_{8,8}.(0.7)^{8}.(0.3)^{0} = 0.0576[/tex]
Then:
[tex]P(X \geq 7) = P(X = 7) + P(X = 8) = 0.1977 + 0.0576 = 0.2553[/tex]
[tec]P(X < 7) = 1 - P(X \geq 7) = 1 - 0.2553 = 0.7447[/tex]
0.7447 = 74.47% probability that fewer than 7 of them show up.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
#SPJ1
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.
