Chapter 5.2, Problem 26P

a.

To determine

Obtain a trial for the given problem.

Describe a success for the given problem.

Describe a failure for the given problem.

Mention the values for n, p, and q.

Compute the probability that at least half the patients are under the age of 15 years and explain the way to model the given problem as binomial distribution.

Answer to Problem 26P

The trial is visit to office.

The success is described as visiting patient’s age is under 15 years.

The failure is described as visiting patient’s age is 15 years or older.

Total number of office visits is $n=8$.

The probability that visiting patient’s age is under 15 years (or success) is $p=0.20$.

The probability that visiting patient’s age is 15 years or older is $q=0.80$.

The probability that at least half the patients are under the age of 15 years is 0.056.

Explanation of Solution

Calculation:

Trial:

The trial is considered as visiting to office.

Success:

Consider success as visiting patient’s age is less than 15 years.

Failure:

Consider failure as visiting patient’s age is 15 years or older.

Values:

Total number of office visits is $n=8$.

The probability that visiting patient’s age is under 15 years (or success) is $p=0.20$.

The probability that visiting patient’s age is 15 years or older (or failure) is calculated as given below:

$\begin{array}{c}q=1-p\\ =1-0.20\\ =0.80\end{array}$

Random Variable:

Let r be a binomial random variable, which represents the number of patients under the age of 15 years.

Binomial probability:

The probability of r successes out of n trials is given below:

$P(r)=C_{n,r}{p}^r{q}^{n-r}$

Here, n is the number of trials, r is the number of successes, p is the probability of success, and q is the probability of failure.

Here, there are only two possible outcomes, the trials are independent of each other, the probability that visiting patient’s age is under 15 years remains same for all the trials, and there are fixed number of trials (n=8). Thus, this can be modeled as binomial distribution.

The probability that at least half the patients are under the age of 15 years is calculated as given below:

$\begin{array}{c}P\left(r\ge 4\right)=P\left(4\right)+P\left(5\right)+P\left(6\right)+P\left(7\right)+P\left(8\right)\\ =C_{8,4}{\left(0.20\right)}^4{\left(0.80\right)}^4+C_{8,5}{\left(0.20\right)}^5{\left(0.80\right)}^3+C_{8,6}{\left(0.20\right)}^6{\left(0.80\right)}^2+C_{8,7}{\left(0.20\right)}^7{\left(0.80\right)}^1\\ +C_{8,8}{\left(0.20\right)}^8{\left(0.80\right)}^0\\ =0.046+0.009+0.001+0.000+0.000\\ =0.056\end{array}$

Therefore, the probability that at least half the patients are under the age of 15 years is 0.056.

b.

To determine

Compute the probability that 2 to 5 patients are 65 years or older.

Answer to Problem 26P

The probability that 2 to 5 patients are 65 years or older are 0.629

Explanation of Solution

Calculation:

Here, the success is visiting patient’s age is 65 years or older and the failure is visiting patient’s age is less than 65 years.

The probability that visiting patient’s age is 65 years or older (or success) is $p=0.25$

The probability that visiting patient’s age less than 65 years is calculated as given below:

$\begin{array}{c}q=1-p\\ =1-0.25\\ =0.75\end{array}$

The probability that 2 to 5 patients are 65 years or older is calculated as given below:

$\begin{array}{c}P\left(2\le r\le 5\right)=P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)\\ =C_{8,2}{\left(0.25\right)}^2{\left(0.75\right)}^6+C_{8,3}{\left(0.25\right)}^3{\left(0.75\right)}^5+C_{8,4}{\left(0.25\right)}^4{\left(0.75\right)}^4\\ +C_{8,5}{\left(0.25\right)}^5{\left(0.75\right)}^3\\ =0.311+0.208+0.087+0.023\\ =0.629\end{array}$

Therefore, the probability that 2 to 5 patients are 65 years or older is 0.629.

c.

To determine

Compute the probability that 2 to 5 patients are 45 years or older.

Answer to Problem 26P

The probability that 2 to 5 patients are 45 years or older is 0.849.

Explanation of Solution

Calculation:

Here, the success is visiting patient’s age is 45 years or older and the failure is visiting patient’s age is less than 45 years.

The probability that visiting patient’s age is 45 years or older (or success) is calculated as given below:

$\begin{array}{c}p=45\text{ years old or older}\\ =0.20+0.25\left(\text{From given table}\right)\\ =0.45\end{array}$

The probability that visiting patient’s age less than 45 years (or failure) is calculated as given below:

$\begin{array}{c}q=1-p\\ =1-0.45\\ =0.55\end{array}$

Random Variable:

Let r be a binomial random variable, which represents the number of patients 45 years or older.

The probability that 2 to 5 patients are 45 years or older is calculated as given below:

$\begin{array}{c}P\left(2\le r\le 5\right)=P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)\\ =C_{8,2}{\left(0.45\right)}^2{\left(0.55\right)}^6+C_{8,3}{\left(0.45\right)}^3{\left(0.55\right)}^5+C_{8,4}{\left(0.45\right)}^4{\left(0.55\right)}^4\\ +C_{8,5}{\left(0.45\right)}^5{\left(0.55\right)}^3\\ =0.157+0.257+0.263+0.172\\ =0.849\end{array}$

Therefore, the probability that 2 to 5 patients are 45 years or older is 0.849.

d.

To determine

Compute the probability that all the patients are 25 years old.

Answer to Problem 26P

The probability that all the patients are less than 25 years or older is 0.000.

Explanation of Solution

Calculation:

Here, the success is visiting patient’s age is less than 25 years and the failure is visiting patient’s age is 25 years or older.

The probability that visiting patient’s age is less than 25 years (or success) is calculated as given below:

$\begin{array}{c}p=\text{under 15 years }+\text{ under}\left(15-24\right)\\ =0.20+0.10\left(\text{From given table}\right)\\ =0.30\end{array}$

The probability that visiting patient’s age is 25 years or older (or failure) is calculated as given below:

$\begin{array}{c}q=1-p\\ =1-0.30\\ =0.70\end{array}$

Random Variable:

Let r be a binomial random variable, which represents the number of patients under 25 years.

The probability that all the patients are less than 25 years is calculated as given below:

$\begin{array}{c}P\left(8\right)=C_{8,8}{\left(0.30\right)}^8{\left(0.70\right)}^0\\ =1\times {\left(0.30\right)}^8\times 1\\ =0.000\end{array}$

Therefore, the probability that all the patients are less than 25 years is 0.000.

e.

To determine

Compute the probability that all the patients are 15 years old or older.

Answer to Problem 26P

The probability that all the patients are 15 years old or older is 0.168.

Explanation of Solution

Calculation:

Here, the success is visiting patient’s age is 15 years old or older and the failure is visiting patient’s age is less than 15 years.

The probability that visiting patient’s age is 15 years old or older (or success) is calculated as given below:

$\begin{array}{c}p=\text{under}\left(15-24\right)+\left(25-44\right)+\left(45-64\right)+65\text{ and older}\\ =0.10+0.25+0.20+0.25\left(\text{From given table}\right)\\ =0.80\end{array}$

The probability that visiting patient’s age is under 15 years old (or failure) is calculated as given below:

$\begin{array}{c}q=1-p\\ =1-0.80\\ =0.20\end{array}$

Random Variable:

Let r be a binomial random variable, which represents the number of patients 15 years old or older.

The probability that all the patients are 15 years old or older is calculated as given below:

$\begin{array}{c}P\left(8\right)=C_{8,8}{\left(0.80\right)}^8{\left(0.20\right)}^0\\ =1\times {\left(0.80\right)}^8\times 1\\ =0.168\end{array}$

Therefore, the probability that all the patients are 15 years old or older is 0.168.