Assume that a procedure yields a binomial distribution with a trial repeated

usagirl007A 2021-10-01 Answered

Assume that a procedure yields a binomial distribution with a trial repeated \(\displaystyle{n}={5}\) times. Use some form of technology to find the cumulative probability distribution given the probability \(\displaystyle{p}={0.16}\) of success on a single trial.
(Report answers accurate to 4 decimal places.)
\(\begin{array}{|c|c|}\hline k & P(X=k) \\ \hline 0 & \\ \hline 1 & \\ \hline 2 & \\ \hline 3 & \\ \hline 4 & \\ \hline 5 & \\ \hline \end{array}\)

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Expert Answer

ottcomn
Answered 2021-10-02 Author has 12767 answers

Step 1
The binomial probability distribution is,
\(P(X=k)=\left(\begin{array}{c}n\\ x\end{array}\right)(p)^{k}(1-p)^{n-k}\)
In the formula, n denotes the number of trails, p denotes probability of success, and k denotes the number of success.
Step 2
The probability of \(\displaystyle{k}={0}\) is,
\(P(X=0)=\left(\begin{array}{c}5\\ 0\end{array}\right)(0.16)^{0}(1-0.16)^{5-0}\)
\(\displaystyle={0.4182}\)
The probability of \(\displaystyle{k}={1}\) is,
\(P(X=1)=\left(\begin{array}{c}5\\ 1\end{array}\right)(0.16)^{1}(1-0.16)^{5-1}\)
\(\displaystyle={0.3983}\)
The probability of \(\displaystyle{k}={2}\) is,
\(P(X=2)=\left(\begin{array}{c}5\\ 2\end{array}\right)(0.16)^{2}(1-0.16)^{5-2}\)
\(\displaystyle={0.1517}\)
The probability of \(\displaystyle{k}={3}\) is,
\(P(X=3)=\left(\begin{array}{c}5\\ 3\end{array}\right)(0.16)^{3}(1-0.16)^{5-3}\)
\(\displaystyle={0.0289}\)
The probability of \(\displaystyle{k}={4}\) is,
\(P(X=4)=\left(\begin{array}{c}5\\ 4\end{array}\right)(0.16)^{4}(1-0.16)^{5-4}\)
\(\displaystyle={0.0028}\)
The probability of \(\displaystyle{k}={5}\) is,
\(P(X=5)=\left(\begin{array}{c}5\\ 5\end{array}\right)(0.16)^{5}(1-0.16)^{5-5}\)
\(\displaystyle={0.0001}\)
Thus, the probability distribution given the probability \(\displaystyle{p}={0.16}\) of success on a single trial is,
\(\begin{array}{|c|c|} \hline k & P(X=k) \\ \hline 0 & 0.4182\\ \hline 1&0.3983\\ \hline2&0.1517\\ \hline3&0.0289\\ \hline4&0.0028\\ \hline5&0.0001 \\ \hline \end{array}\)

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