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Sagot :
Using the binomial distribution, it is found that there is a:
- 0.2965 = 29.65% probability that exactly 2 out of 8 randomly selected Americans own a cat.
- 0.4482 = 44.82% probability that more than 3 own a cat.
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, we have that:
- 30% of Americans own a cat, hence p = 0.3.
- A sample of 8 Americans is taken, hence n = 8.
The probability that exactly 2 out of 8 randomly selected Americans own a cat is P(X = 2), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{8,2}.(0.3)^{2}.(0.7)^{6} = 0.2965[/tex]
0.2965 = 29.65% probability that exactly 2 out of 8 randomly selected Americans own a cat.
The probability that more than 3 own a cat is given by:
[tex]P(X > 3) = 1 - P(X \leq 2)[/tex]
In which:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
Hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{8,0}.(0.3)^{0}.(0.7)^{8} = 0.0576[/tex]
[tex]P(X = 1) = C_{8,1}.(0.3)^{1}.(0.7)^{7} = 0.1977[/tex]
[tex]P(X = 2) = C_{8,2}.(0.3)^{2}.(0.7)^{6} = 0.2965[/tex]
Then:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0576 + 0.1977 + 0.2965 = 0.5518[/tex]
[tex]P(X > 3) = 1 - P(X \leq 2) = 1 - 0.5518 = 0.4482[/tex]
0.4482 = 44.82% probability that more than 3 own a cat.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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