Respuesta :
Answer: a) 0.17, b) 0.0363
Step-by-step explanation:
Since we have given that
Number of jurors selected = 12
Number of minorities of jurors = 2
(a) What proportion of the jury described is from a minority race?
[tex]\dfrac{2}{12}=\dfrac{1}{6}=0.166666\approx0.17[/tex]
(b) If 12 jurors are randomly selected from a population where 46% are minorities, what is the probability that 2 or fewer jurors will be minorities?
Let X be the number of jurors who are minorities.
Probability of getting minorities = 0.46
Probability of getting no minorities = 1-0.46 =0.54
[tex]P(X\leq 2)=^{12}C_0(0.54)^{12}+^{12}C_1(0.46)(0.54)^{11}+^{12}C_2(0.46)^2(0.54)^{10}\\\\P(X\leq 2)=0.0363[/tex]
(c) What might the lawyer of a defendant from this minority race argue?
Since the number of minorities juror are high among the population.
Hence, a) 0.17, b) 0.0363
Using the binomial distribution, it is found that:
a) 0.1667 = 16.67% of the jury described is from a minority race.
b) 0.0363 = 3.63% probability that 2 or fewer jurors will be minorities.
c) There is a low probability of a jury with 2 or fewer jurors as minorities, thus, the lawyer could argue that the lack of minorities in the jury was done on purpose to harm his client.
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For each juror, there are only two possible outcomes. Either it is a minority, or it is not. The probability of a juror being a minority is independent of any other juror, 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, which is the number of successes.
- n, which is the number of trials.
- p, which is the probability of a success on a single trial.
In this problem:
- 12 jurors, thus [tex]n = 12[/tex].
- 46% minorities, thus [tex]p = 0.46[/tex].
Item a:
2 out of 12 are minorities, thus:
[tex]p = \frac{2}{12} = \frac{1}{6} = 0.1667[/tex]
0.1667 = 16.67% of the jury described is from a minority race.
Item b:
This is:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{12,0}.(0.46)^{0}.(0.54)^{12} = 0.0006[/tex]
[tex]P(X = 1) = C_{12,1}.(0.46)^{1}.(0.54)^{11} = 0.0063[/tex]
[tex]P(X = 2) = C_{12,2}.(0.46)^{2}.(0.54)^{10} = 0.0294[/tex]
Then
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0006 + 0.0063 + 0.0294 = 0.0363[/tex]
0.0363 = 3.63% probability that 2 or fewer jurors will be minorities.
Item c:
There is a low probability of a jury with 2 or fewer jurors as minorities, thus, the lawyer could argue that the lack of minorities on the jury was done on purpose to harm his client.
A similar problem is given at https://brainly.com/question/13965196
