Question

coronary bypass surgery: a healthcare agency reported that 49% of people who had coronary bypass surgery in a recent year were over the age of 65. fifteen coronary bypass patients are sampled.
part 1 of 4
(a) what is the probability that exactly 10 of them are over the age of 65? round the answer to four decimal places.
the probability that exactly 10 of them are over the age of 65 is 0.0827
part 1 / 4
part 2 of 4
(b) what is the probability that more than 11 are over the age of 65? round the answer to four decimal places.
the probability that more than 11 are over the age of 65 is

Explanation:

Step 1: Identify the 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 \(C(n,k)=\frac{n!}{k!(n - k)!}\)

For part (a), \(n = 15\), \(k = 10\), \(p=0.49\), \(1-p = 0.51\)

First, calculate \(C(15,10)=\frac{15!}{10!(15 - 10)!}=\frac{15!}{10!5!}=\frac{15\times14\times13\times12\times11}{5\times4\times3\times2\times1}=3003\)

Then \(P(X = 10)=3003\times(0.49)^{10}\times(0.51)^{5}\)

\((0.49)^{10}\approx0.000282475\), \((0.51)^{5}\approx0.034502525\)

\(P(X = 10)=3003\times0.000282475\times0.034502525\approx0.0287\)

For part (b), \(P(X>11)=P(X = 12)+P(X = 13)+P(X = 14)+P(X = 15)\)

\(C(15,12)=\frac{15!}{12!(15 - 12)!}=\frac{15\times14\times13}{3\times2\times1}=455\)

\(P(X = 12)=455\times(0.49)^{12}\times(0.51)^{3}\)

\((0.49)^{12}\approx0.00001342\), \((0.51)^{3}\approx0.132651\)

\(P(X = 12)=455\times0.00001342\times0.132651\approx0.0008\)

\(C(15,13)=\frac{15!}{13!(15 - 13)!}=\frac{15\times14}{2\times1}=105\)

\(P(X = 13)=105\times(0.49)^{13}\times(0.51)^{2}\)

\((0.49)^{13}\approx0.00000657\), \((0.51)^{2}=0.2601\)

\(P(X = 13)=105\times0.00000657\times0.2601\approx0.0002\)

\(C(15,14)=\frac{15!}{14!(15 - 14)!}=15\)

\(P(X = 14)=15\times(0.49)^{14}\times(0.51)^{1}\)

\((0.49)^{14}\approx0.00000322\), \((0.51)^{1}=0.51\)

\(P(X = 14)=15\times0.00000322\times0.51\approx0.000025\)

\(C(15,15)=\frac{15!}{15!(15 - 15)!}=1\)

\(P(X = 15)=(0.49)^{15}\approx0.00000158\)

\(P(X>11)=0.0008 + 0.0002+0.000025 + 0.00000158\approx0.0010\)

Answer:

(a) \(0.0287\)
(b) \(0.0010\)