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assume that a procedure yields a binomial distribution with a trial rep…

Question

assume that a procedure yields a binomial distribution with a trial repeated \\(n = 5\\) times. use some form of technology to find the probability distribution given the probability \\(p = 0.234\\) of success on a single trial.

(report answers accurate to 4 decimal places.)

\\(\

$$\begin{array}{|c|c|} \\hline k & p(x = k) \\\\ \\hline 0 & \\\\ \\hline 1 & \\\\ \\hline 2 & .2461 \\\\ \\hline 3 & .0752 \\\\ \\hline 4 & .0115 \\\\ \\hline 5 & .0007 \\\\ \\hline \\end{array}$$

\\)

Explanation:

Identify given parameters

We are given a binomial distribution with:

  • Number of trials: \(n = 5\)
  • Probability of success: \(p = 0.234\)
  • Probability of failure: \(q = 1 - p = 1 - 0.234 = 0.766\)

State the binomial formula

The probability of getting exactly \(k\) successes is:

$$P(X = k) = \binom{n}{k} p^k q^{n-k}$$

Calculate for k = 0

Substitute \(k = 0\) into the formula:

$$P(X = 0) = \binom{5}{0} (0.234)^0 (0.766)^5$$
$$P(X = 0) = 1 \times 1 \times (0.766)^5 \approx 0.266388$$

Rounding to 4 decimal places:

$$P(X = 0) \approx 0.2664$$

Calculate for k = 1

Substitute \(k = 1\) into the formula:

$$P(X = 1) = \binom{5}{1} (0.234)^1 (0.766)^4$$
$$P(X = 1) = 5 \times 0.234 \times (0.766)^4$$
$$P(X = 1) = 1.17 \times 0.344331 \approx 0.402867$$

Rounding to 4 decimal places:

$$P(X = 1) \approx 0.4029$$

Answer:

Assume that a procedure yields a binomial distribution with a trial repeated \(n = 5\) times. Use some form of technology to find the probability distribution given the probability \(p = 0.234\) of success on a single trial.

(Report answers accurate to 4 decimal places.)

  • For \(k = 0\), \(P(X = 0) =\) <blank>0.2664</blank>
  • For \(k = 1\), \(P(X = 1) =\) <blank>0.4029</blank>