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

Question

Accepted Answer

# Explanation:
## Step1: Identify the distribution
This is a binomial distribution problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 15$ (number of trials/sampled patients), $p=0.49$ (probability of success/patient over 65), and $C(n,k)=\frac{n!}{k!(n - k)!}$

## Step2: Calculate for $k = 12$
$$
\begin{align*}
C(15,12)&=\frac{15!}{12!(15 - 12)!}=\frac{15\times14\times13\times12!}{12!\times3!}=\frac{15\times14\times13}{3\times2\times1}=455\
P(X = 12)&=C(15,12)\times(0.49)^{12}\times(1 - 0.49)^{15 - 12}\
&=455\times(0.49)^{12}\times(0.51)^{3}
\end{align*}
$$
$$
\begin{align*}
(0.49)^{12}&\approx0.00061\
(0.51)^{3}&=0.51\times0.51\times0.51 = 0.132651\
P(X = 12)&=455\times0.00061\times0.132651\approx0.0366
\end{align*}
$$

## Step3: Calculate for $k = 13$
$$
\begin{align*}
C(15,13)&=\frac{15!}{13!(15 - 13)!}=\frac{15\times14\times13!}{13!\times2!}=\frac{15\times14}{2\times1}=105\
P(X = 13)&=C(15,13)\times(0.49)^{13}\times(0.51)^{2}\
(0.49)^{13}&\approx0.0003\
(0.51)^{2}&=0.2601\
P(X = 13)&=105\times0.0003\times0.2601\approx0.0082
\end{align*}
$$

## Step4: Calculate for $k = 14$
$$
\begin{align*}
C(15,14)&=\frac{15!}{14!(15 - 14)!}=\frac{15\times14!}{14!\times1!}=15\
P(X = 14)&=C(15,14)\times(0.49)^{14}\times(0.51)^{1}\
(0.49)^{14}&\approx0.00015\
P(X = 14)&=15\times0.00015\times0.51\approx0.0011
\end{align*}
$$

## Step5: Calculate for $k = 15$
$$
\begin{align*}
C(15,15)&=\frac{15!}{15!(15 - 15)!}=1\
P(X = 15)&=(0.49)^{15}\approx0.00007
\end{align*}
$$

## Step6: Sum the probabilities
$P(X>11)=P(X = 12)+P(X = 13)+P(X = 14)+P(X = 15)$
$P(X>11)=0.0366 + 0.0082+0.0011 + 0.00007\approx0.0459$

# Answer:
$0.0459$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the distribution

Step2: Calculate for \(k = 12\)

Step3: Calculate for \(k = 13\)

Step4: Calculate for \(k = 14\)

Step5: Calculate for \(k = 15\)

Step6: Sum the probabilities

Answer: