Respuesta :
Using the binomial distribution, it is found that there is a 0.0555 = 5.55% probability that more than 3 of these selected children have three siblings.
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, for the parameters, we consider that:
- 10% of the 9-year old children in a town have three siblings, hence p = 0.1.
- Fifteen 9-year old children are selected at random, hence n = 15.
The probability that more than 3 of these selected children have three siblings is given by:
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which:
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059[/tex]
[tex]P(X = 1) = C_{15,1}.(0.1)^{1}.(0.9)^{14} = 0.3432[/tex]
[tex]P(X = 2) = C_{15,2}.(0.1)^{2}.(0.9)^{13} = 0.2669[/tex]
[tex]P(X = 3) = C_{15,3}.(0.1)^{3}.(0.9)^{12} = 0.1285[/tex]
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.2059 + 0.3432 + 0.2669 + 0.1285 = 0.9445[/tex]
P(X > 3) = 1 - 0.9445 = 0.0555
0.0555 = 5.55% probability that more than 3 of these selected children have three siblings.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
