Chapter 5.4, Problem 32P

a.

To determine

Explain the reason for a negative binomial distribution to be appropriate for a given random variable and define a formula for $P\left(n\right)$ in the given context of application.

Answer to Problem 32P

There are binomial trials with probability of success 0.41 and failure 0.59 with n as a random variable that represents the number of donors required to provide 3 pints of type A blood.

The formula for probability that a clinic needs 3 pints of type A blood is. $\underset{\_}{P\left(n\right)=C_{n-1,2}{\left(0.41\right)}^3{\left(0.59\right)}^{n-3}}$

Explanation of Solution

Calculation:

There are binomial trials with probability of success $p=0.41$ and failure as $q=0.59$.

Let $n$ follow a negative binomial distribution that represents the number of donors required to provide 3 pints of type A blood.

Probability of success $p=0.41$

Probability of failure $q=0.59$

Number of pints $k=3$

Negative binomial probability:

The probability that k^th success occurs on n^th trial is given below:

$P\left(n\right)=C_{n-1,k-1}{p}^k{q}^{n-k}$

Here, $n$ is the number of trials in which k^th success occurs, $k$ is the number of successes, $p$ is the probability of success, and $q$ is the probability of failure.

The formula for probability that a clinic needs 3 pints of type A blood is given below:

$\begin{array}{l}P\left(n\right)=C_{n-1,k-1}{p}^k{q}^{n-k}\\ =C_{n-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{n-3}\\ =C_{n-1,2}{\left(0.41\right)}^3{\left(0.59\right)}^{n-3}\end{array}$

Thus, the formula for probability that a clinic needs 3 pints of type A blood is. $\underset{\_}{P\left(n\right)=C_{n-1,2}{\left(0.41\right)}^3{\left(0.59\right)}^{n-3}}$

b.

To determine

Calculate the given probabilities.

Answer to Problem 32P

The probability that a clinic needs 3 donors is 0.0689.

The probability that a clinic needs 4 donors is 0.1220.

The probability that a clinic needs 5 donors is 0.1439.

The probability that a clinic needs 6 donors is 0.1415.

Explanation of Solution

Calculation:

The probability that a clinic needs 3 donors is given below:

$\begin{array}{l}P\left(3\right)=C_{3-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{3-3}\\ =C_{2,2}{\left(0.41\right)}^3{\left(0.59\right)}^0\\ =\left(\frac{2!}{2!0!}\right){\left(0.41\right)}^3{\left(0.59\right)}^0\\ =0.0689\end{array}$

Thus, the probability that a clinic needs 3 donors is 0.0689.

The probability that a clinic needs 4 donors is given below:

$\begin{array}{l}P\left(4\right)=C_{4-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{4-3}\\ =C_{3,2}{\left(0.41\right)}^3{\left(0.59\right)}^1\\ =\left(\frac{3!}{2!1!}\right){\left(0.41\right)}^3{\left(0.59\right)}^1\\ =0.1220\end{array}$

Thus, the probability that a clinic needs 4 donors is 0.1220.

The probability that a clinic needs 5 donors is given below:

$\begin{array}{l}P\left(5\right)=C_{5-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{5-3}\\ =C_{4,2}{\left(0.41\right)}^3{\left(0.59\right)}^2\\ =\left(\frac{4!}{2!2!}\right){\left(0.41\right)}^3{\left(0.59\right)}^2\\ =0.1439\end{array}$

Thus, the probability that a clinic needs 5 donors is 0.1439.

The probability that a clinic needs 6 donors is given below:

$\begin{array}{l}P\left(6\right)=C_{6-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{6-3}\\ =C_{5,2}{\left(0.41\right)}^3{\left(0.59\right)}^3\\ =\left(\frac{5!}{2!3!}\right){\left(0.41\right)}^3{\left(0.59\right)}^3\\ =0.1415\end{array}$

Thus, the probability that a clinic needs 6 donors is 0.1415.

c.

To determine

Calculate the probability that a clinic will need 3 to 6 donors to obtain 3 pints of type A blood.

Answer to Problem 32P

The probability that a clinic will need 3 to 6 donors to obtain 3 pints of type A blood.

Explanation of Solution

Calculation:

The probability that a clinic will need 3 to 6 donors to obtain 3 pints of type A blood is calculated as given below:

$\begin{array}{l}P\left(3\le n\le 6\right)=P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)\\ =C_{3-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{3-3}+C_{4-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{4-3}+C_{5-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{5-3}\\ +C_{6-1,3-1}{\left(0.41\right)}^3{\left(0.59\right)}^{6-3}\\ =0.0689+0.1220+0.1439+0.1415\\ =\text{0}\text{.4763}\end{array}$

Thus, the probability that a clinic will need 3 to 6 donors to obtain 3 pints of type A blood is 0.4763.

d.

To determine

Calculate the probability that a clinic will need more than 6 donors to obtain 3 pints of type A blood.

Answer to Problem 32P

The probability that a clinic will need more than 6 donors to obtain 3 pints of type A is 0.5237.

Explanation of Solution

Calculation:

The probability that a clinic will need more than 6 donors to obtain 3 pints of type A is calculated as given below:

$\begin{array}{l}P\left(n>6\right)=1-P\left(n\le 6\right)\\ =1-P\left(3\le n\le 6\right)\\ =1-\text{0}\text{.4763}\\ =\text{0}\text{.5237}\end{array}$

Thus, the probability that a clinic will need more than 6 donors to obtain 3 pints of type A is 0.5237.

e.

To determine

Calculate the expected value of $n$.

Calculate the standard deviation of n.

Answer to Problem 32P

The expected value of n is 7.32

The standard deviation of n is 3.24

Explanation of Solution

Calculation:

The expected value of n is calculated as follows:

$\begin{array}{l}\mu =\frac{k}{p}\\ =\frac{3}{0.41}\\ =7.32\end{array}$

Thus, the expected value of n is 7.32.

The standard deviation of n is calculated as follows:

$\begin{array}{l}\sigma =\frac{\sqrt{kq}}{p}\\ =\frac{\sqrt{3\times 0.59}}{0.41}\\ =3.24\end{array}$

Thus, the standard deviation of n is 3.24.

Interpretation:

The expected number of donors that 3 pints of blood type A is obtained is 7.32 with a standard deviation of 3.24.