assume that a procedure yields a binomial distribution with a trial repeated n = 9 times. use either the…

assume that a procedure yields a binomial distribution with a trial repeated n = 9 times. use either the binomial probability formula (or a technology like excel or statdisk) to find the probability of k = 6 successes given the probability p =3/10 of success on a single trial. (report answer accurate to 4 decimal places.) p(x = k) =

assume that a procedure yields a binomial distribution with a trial repeated n = 9 times. use either the binomial probability formula (or a technology like excel or statdisk) to find the probability of k = 6 successes given the probability p =3/10 of success on a single trial. (report answer accurate to 4 decimal places.) p(x = k) =

Answer

Explanation:

Step1: Recall binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Calculate the combination $C(n,k)$

Given $n = 9$ and $k = 6$. $C(9,6)=\frac{9!}{6!(9 - 6)!}=\frac{9!}{6!3!}=\frac{9\times8\times7}{3\times2\times1}=84$.

Step3: Calculate $(1 - p)$

Given $p=\frac{3}{10}$, then $1 - p=1-\frac{3}{10}=\frac{7}{10}$.

Step4: Substitute values into the binomial - probability formula

$P(X = 6)=C(9,6)\times p^{6}\times(1 - p)^{9 - 6}$ $=84\times(\frac{3}{10})^{6}\times(\frac{7}{10})^{3}$ $=84\times\frac{729}{1000000}\times\frac{343}{1000}$ $=84\times\frac{729\times343}{1000000\times1000}$ $=84\times\frac{249047}{1000000000}$ $=\frac{20920948}{1000000000}\approx0.0209$.

Answer:

$0.0209$