Respuesta :
Answer:
Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.
Step-by-step explanation:
We are given that an annexation suit against a county subdivision of 1200 residences is being considered by a neighboring city. The occupants of half the residences object to being annexed.
Also, a random sample of 10 residents is taken.
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 = 10 residents
r = number of success = at least 3
p = probability of success which in our question is probability that
residents favor the annexation suit, which is calculated as below;
p = [tex]\frac{\text{Number of residents favoring the annexation suit }}{\text{Total number of residents considered } }[/tex] = [tex]\frac{600}{1200}[/tex] = 0.50
LET X = Number of residents favoring the annexation suit
So, it means X ~ [tex]Binom(n=10, p=0.50)[/tex]
Now, Probability that in a random sample of 10 at least 3 favor the annexation suit is given by = P(X [tex]\geq[/tex] 3)
P(X [tex]\geq[/tex] 3) = 1 - P(X < 3) = 1 - P(X [tex]\leq[/tex] 2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= [tex]1- [\binom{10}{0}\times 0.50^{0} \times (1-0.50)^{10-0} + \binom{10}{1}\times 0.50^{1} \times (1-0.50)^{10-1} +\binom{10}{2}\times 0.50^{2} \times (1-0.50)^{10-2}][/tex]
= [tex]1-[ 1 \times 1 \times 0.50^{10}+10 \times 0.50^{1} \times 0.50^{9}+45 \times 0.50^{2} \times 0.50^{8}][/tex]
= [tex]1-[ 0.50^{10}+10 \times 0.50^{10}+45 \times 0.50^{10}][/tex]
= [tex]1-0.50^{10}[ 1+10 +45 ][/tex] = [tex]1-0.50^{10} \times 56[/tex]
= 0.9453
Therefore, Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.
Using the binomial distribution, it is found that there is a 0.9453 = 94.53% probability that in a random sample of 10 at least 3 favor the annexation suit.
For each resident, there are only two possible outcomes, either they favor the suit, or they do not. The probability of a resident favoring the suit is independent of any other resident, hence the binomial distribution is used to solve this question.
What is the binomial probability 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:
- Half favor the annexation suit, hence [tex]p = 0.5[/tex].
- A sample of 10 is taken, hence [tex]n = 10[/tex].
The probability that in a random sample of 10 at least 3 favor the annexation suit is:
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.5)^{0}.(0.5)^{10} = 0.0010[/tex]
[tex]P(X = 1) = C_{10,1}.(0.5)^{1}.(0.5)^{9} = 0.0098[/tex]
[tex]P(X = 2) = C_{10,2}.(0.5)^{2}.(0.5)^{8} = 0.0439[/tex]
Then:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0010 + 0.0098 + 0.0439 = 0.0547[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.0547 = 0.9453[/tex]
0.9453 = 94.53% probability that in a random sample of 10 at least 3 favor the annexation suit.
You can learn more about the binomial distribution at https://brainly.com/question/24863377
