Resolved: "Assume that a procedure yields a binomial distribution..."

usagirl007A

Answered question

2021-10-01

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

Answer & Explanation

ottcomn

Skilled2021-10-02Added 97 answers

Step 1
The binomial probability distribution is,
$P (X = k) = (\begin{matrix} n \\ x \end{matrix}) (p)^{k} (1 - p)^{n - k}$
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 $k = 0$ is,
$P (X = 0) = (\begin{matrix} 5 \\ 0 \end{matrix}) (0.16)^{0} (1 - 0.16)^{5 - 0}$
$= 0.4182$
The probability of $k = 1$ is,
$P (X = 1) = (\begin{matrix} 5 \\ 1 \end{matrix}) (0.16)^{1} (1 - 0.16)^{5 - 1}$
$= 0.3983$
The probability of $k = 2$ is,
$P (X = 2) = (\begin{matrix} 5 \\ 2 \end{matrix}) (0.16)^{2} (1 - 0.16)^{5 - 2}$
$= 0.1517$
The probability of $k = 3$ is,
$P (X = 3) = (\begin{matrix} 5 \\ 3 \end{matrix}) (0.16)^{3} (1 - 0.16)^{5 - 3}$
$= 0.0289$
The probability of $k = 4$ is,
$P (X = 4) = (\begin{matrix} 5 \\ 4 \end{matrix}) (0.16)^{4} (1 - 0.16)^{5 - 4}$
$= 0.0028$
The probability of $k = 5$ is,
$P (X = 5) = (\begin{matrix} 5 \\ 5 \end{matrix}) (0.16)^{5} (1 - 0.16)^{5 - 5}$
$= 0.0001$
Thus, the probability distribution given the probability $p = 0.16$ of success on a single trial is,
$\begin{array}{cc} k & P (X = k) \\ 0 & 0.4182 \\ 1 & 0.3983 \\ 2 & 0.1517 \\ 3 & 0.0289 \\ 4 & 0.0028 \\ 5 & 0.0001 \end{array}$

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