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 = 0.527 ) of success on a single trial.
(report answers accurate to 4 decimal places.)

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 \(n = 5\), \(p=0.527\), and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate \(C(n,k)\) for each \(k\)

• For \(k = 0\):

\(C(5,0)=\frac{5!}{0!(5-0)!}=1\)
\(P(X = 0)=1\times(0.527)^{0}\times(1 - 0.527)^{5-0}=1\times1\times(0.473)^{5}\approx0.0222\)

• For \(k = 1\):

\(C(5,1)=\frac{5!}{1!(5 - 1)!}=\frac{5!}{1!4!}=5\)
\(P(X = 1)=5\times(0.527)^{1}\times(0.473)^{4}\)
\(=5\times0.527\times0.0473^{4}\approx0.1311\)

• For \(k = 2\):

\(C(5,2)=\frac{5!}{2!(5-2)!}=\frac{5\times4}{2\times1}=10\)
\(P(X = 2)=10\times(0.527)^{2}\times(0.473)^{3}\)
\(=10\times0.277729\times0.105823\approx0.2939\)

• For \(k = 3\):

\(C(5,3)=\frac{5!}{3!(5 - 3)!}=\frac{5\times4}{2\times1}=10\)
\(P(X = 3)=10\times(0.527)^{3}\times(0.473)^{2}\)
\(=10\times0.1477\times0.2237\approx0.3274\)

• For \(k = 4\):

\(C(5,4)=\frac{5!}{4!(5-4)!}=5\)
\(P(X = 4)=5\times(0.527)^{4}\times(0.473)^{1}\)
\(=5\times0.0779\times0.473\approx0.1822\)

• For \(k = 5\):

\(C(5,5)=\frac{5!}{5!(5 - 5)!}=1\)
\(P(X = 5)=1\times(0.527)^{5}\times(0.473)^{0}\)
\(=(0.527)^{5}\approx0.0406\)

Answer:

The probabilities \(P(X = k)\) for \(k = 0,1,2,3,4,5\) are \(0.0222\), \(0.1311\), \(0.2939\), \(0.3274\), \(0.1822\), \(0.0406\) respectively.