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Sagot :
Using the binomial distribution, it is found that there is a:
a) 0.2573 = 25.73% probability that the number of U.S. adults who favor the use of unmanned drones by police agencies is exactly three.
b) 0.3205 = 32.05% probability that the number of U.S. adults who favor the use of unmanned drones by police agencies is at least four.
c) 0.9979 = 99.79% probability that the number of U.S. adults who favor the use of unmanned drones by police agencies is less than eight.
For each adult, there are only two possible outcomes. Either they favor the use of drones, or they do not. The probability of an adult favoring the use of drones is independent of any other adult, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[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:
- 24% favor the use of drones, thus [tex]p = 0.24[/tex].
- Sample of 12, thus [tex]n = 12[/tex].
Item a:
- The probability is P(X = 3).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{12,3}.(0.24)^{3}.(0.76)^{9} = 0.2573[/tex]
0.2573 = 25.73% probability that the number of U.S. adults who favor the use of unmanned drones by police agencies is exactly three.
Item b:
The probability is:
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which:
[tex]P(X < 4) = 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_{12,0}.(0.24)^{0}.(0.76)^{12} = 0.0371[/tex]
[tex]P(X = 1) = C_{12,1}.(0.24)^{1}.(0.76)^{11} = 0.1407[/tex]
[tex]P(X = 2) = C_{12,2}.(0.24)^{2}.(0.76)^{10} = 0.2444[/tex]
[tex]P(X = 3) = C_{12,3}.(0.24)^{3}.(0.76)^{9} = 0.2573[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0371 + 0.1407 + 0.2444 + 0.2573 = 0.6795[/tex]
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.9765 = 0.3205[/tex]
0.3205 = 32.05% probability that the number of U.S. adults who favor the use of unmanned drones by police agencies is at least four.
Item c:
This probability is:
[tex]P(X < 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)[/tex]
In which
[tex]P(X = 0) = C_{12,0}.(0.24)^{0}.(0.76)^{12} = 0.0371[/tex]
[tex]P(X = 1) = C_{12,1}.(0.24)^{1}.(0.76)^{11} = 0.1407[/tex]
[tex]P(X = 2) = C_{12,2}.(0.24)^{2}.(0.76)^{10} = 0.2444[/tex]
[tex]P(X = 3) = C_{12,3}.(0.24)^{3}.(0.76)^{9} = 0.2573[/tex]
[tex]P(X = 4) = C_{12,4}.(0.24)^{4}.(0.76)^{8} = 0.1828[/tex]
[tex]P(X = 5) = C_{12,5}.(0.24)^{5}.(0.76)^{7} = 0.0924[/tex]
[tex]P(X = 6) = C_{12,6}.(0.24)^{6}.(0.76)^{6} = 0.0340[/tex]
[tex]P(X = 7) = C_{12,7}.(0.24)^{7}.(0.76)^{5} = 0.0092[/tex]
[tex]P(X < 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0371 + 0.1407 + 0.2444 + 0.2573 + 0.1828 + 0.0924 + 0.0340 + 0.0092 = 0.9979[/tex]
0.9979 = 99.79% probability that the number of U.S. adults who favor the use of unmanned drones by police agencies is less than eight.
A similar problem is given at https://brainly.com/question/24669788
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