Respuesta :
Answer:
Probability that exactly 5 of them favor the building of the health center is 0.0408.
Step-by-step explanation:
We are given that in a recent survey, 60% of the community favored building a health center in their neighborhood.
Also, 14 citizens are chosen.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials (samples) taken = 14 citizens
r = number of success = exactly 5
p = probability of success which in our question is % of the community
favored building a health center in their neighborhood, i.e; 60%
LET X = Number of citizens who favored building of the health center.
So, it means X ~ [tex]Binom(n=14, p=0.60)[/tex]
Now, Probability that exactly 5 of them favor the building of the health center is given by = P(X = 5)
P(X = 5) = [tex]\binom{14}{5} \times 0.60^{5} \times (1-0.60)^{14-5}[/tex]
= [tex]2002 \times 0.60^{5} \times 0.40^{9}[/tex]
= 0.0408
Therefore, Probability that exactly 5 of them favor the building of the health center is 0.0408.
The probability that exactly 5 of the citizens favor the building of the health center is;
P(X = 5) = 0.0408
This question is a binomial probability distribution question that takes the formula;
P(X = x) = ⁿCₓ * pˣ * q⁽ⁿ ⁻ ˣ⁾
Where;
p = chance of success
q = chance of failure
n = number of trials
We are given;
p = 60% = 0.6
q = 1 - 0.6 = 0.4
n = 14
Thus, probability that exactly 5 of them favor the building of the health center is;
P(X = 5) = ¹⁴C₅ × 0.6⁵ × 0.4⁽¹⁴ ⁻ ⁵⁾
P(X = 5) = 2002 × 0.6⁵ × 0.4⁹
P(X = 5) = 0.0408
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