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Suppose you are a lawyer representing a Hispanic individual in a criminal trial. Twelve jurors are randomly selected by the state from a population that you know to be 40% Hispanic.

a. How many jurors would you expect to be Hispanic?


b. The select process produces a jury with 3 Hispanic individuals What proportion of the jury is Hispanic? (Enter your answer as a decimal, rounded to four decimal places)


c. Use the binomial distribution and find the probability that 3 or fewer jurors would be Hispanic.

Respuesta :

Using the binomial distribution, it is found that the desired values are given as follows:

a) 4.8 jurors.

b) The proportion is of 0.25.

c) 0.2254 = 22.54% probability.

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:

  • Twelve jurors are randomly selected by the state, hence n = 12.
  • The state is known to be 40% Hispanic, hence p = 0.4.

Item a:

The expected value of the binomial distribution is given by:

E(X) = np

Hence:

E(X) = 12 x 0.4 = 4.8 jurors.

Item b:

The proportion of 3 out of 12 is 3/12 = 1/4 = 0.25.

Item c:

The probability is given by:

[tex]P(X \leq 3) = 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}.(0.4)^{0}.(0.6)^{12} = 0.0022[/tex]

[tex]P(X = 1) = C_{12,1}.(0.4)^{1}.(0.6)^{11} = 0.0174[/tex]

[tex]P(X = 2) = C_{12,2}.(0.4)^{2}.(0.6)^{10} = 0.0639[/tex]

[tex]P(X = 3) = C_{12,3}.(0.4)^{3}.(0.6)^{9} = 0.1419[/tex]

Then:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0022 + 0.0174 + 0.0639 + 0.1419 = 0.2254[/tex]

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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