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Negative Binomial Distribution: Type A Blood Donors Blood type A occurs in about \(41 \%\) of the population (Reference: Laboratory and Diagnostic Tests by F. Fischbach). A clinic needs 3 pints of type A blood. A donor usually gives a pint of blood. Let \(n\) be a random variable representing the number of donors needed to provide 3 pints of type A blood. (a) Explain why a negative binomial distribution is appropriate for the random variable \(n\). Write out the formula for \(P(n)\) in the context of this application. Hint: See Problem 30 . (b) Compute \(P(n=3), P(n=4), P(n=5)\), and \(P(n=6)\). (c) What is the probability that the clinic will need from three to six donors to obtain the needed 3 pints of type A blood? (d) What is the probability that the clinic will need more than six donors to obtain 3 pints of type A blood? (e) What are the expected value \(\mu\) and standard deviation \(\sigma\) of the random variable \(n\) ? Interpret these values in the context of this application.

Step by step solution

01

Identify the type of distribution

The problem involves finding the probability of obtaining a specific number of successful trials (in this case, obtaining 3 pints of type A blood) when the probability of success (a donor having type A blood) is known. Since each donation is independent, and we're looking for the number of trials to achieve a fixed number of successes, a negative binomial distribution fits this scenario.

02

Write the formula for P(n)

For a negative binomial distribution, the probability mass function is given by:\[P(n = k) = \binom{k-1}{r-1} p^r (1-p)^{k-r}\]where \(r\) is the number of successes (here, 3 pints), \(p\) is the probability of success (0.41), and \(k\) is the total number of donors needed. The expression \(\binom{k-1}{r-1}\) is a binomial coefficient representing the number of ways to arrange \(r-1\) successes in \(k-1\) trials.

03

Calculate P(n=3)

Substitute \(r = 3\), \(p = 0.41\), and \(k = 3\) into the formula:\[P(n = 3) = \binom{2}{2} (0.41)^3 (0.59)^0 = 1 \times 0.06892 \times 1 = 0.06892\]

04

Calculate P(n=4)

Substitute \(k = 4\) into the formula:\[P(n = 4) = \binom{3}{2} (0.41)^3 (0.59)^1 = 3 \times 0.06892 \times 0.59 = 0.12179\]

05

Calculate P(n=5)

Substitute \(k = 5\) into the formula:\[P(n = 5) = \binom{4}{2} (0.41)^3 (0.59)^2 = 6 \times 0.06892 \times 0.3481 = 0.14421\]

06

Calculate P(n=6)

Substitute \(k = 6\) into the formula:\[P(n = 6) = \binom{5}{2} (0.41)^3 (0.59)^3 = 10 \times 0.06892 \times 0.205379 = 0.14111\]

07

Calculate the probability from 3 to 6 donors

Sum the probabilities from Steps 3 to 6:\[P(3 \leq n \leq 6) = P(n=3) + P(n=4) + P(n=5) + P(n=6) = 0.06892 + 0.12179 + 0.14421 + 0.14111 = 0.47603\]

08

Calculate probability of more than 6 donors

The probability of needing more than 6 donors is the complement of needing 6 or fewer donors:\[P(n > 6) = 1 - P(3 \leq n \leq 6) = 1 - 0.47603 = 0.52397\]

09

Calculate expected value and standard deviation

For a negative binomial distribution, the expected value \(\mu\) is given by\[\mu = \frac{r}{p} = \frac{3}{0.41} \approx 7.3171\]The standard deviation \(\sigma\) is \[\sigma = \sqrt{\frac{r(1-p)}{p^2}} = \sqrt{\frac{3(0.59)}{0.41^2}} \approx 2.7077\] The expected value indicates that, on average, about 7.3 donors are needed to get 3 pints, while the standard deviation shows how much the number of donors varies around this average.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Mass Function

In a probability context, the idea of a Probability Mass Function (PMF) becomes particularly pivotal when dealing with discrete random variables, such as the ones found in a Negative Binomial Distribution.

The PMF describes the probability of a certain number of trials being required to achieve a fixed number of successes. In this situation, the variable \( n \) represents the number of donors needed for 3 successful donations of type A blood. Each donor visit is a trial, and each pint of type A blood obtained is a success.

With a Negative Binomial Distribution, the probability mass function is formulated as follows:
\[P(n = k) = \binom{k-1}{r-1} p^r (1-p)^{k-r}\]
Where:

• \( \binom{k-1}{r-1} \) is the binomial coefficient determining the number of ways to arrange \( r-1 \) successes in \( k-1 \) trials.
• \( r \) is the required number of successes (3 pints of blood).
• \( p \) is the probability of success on each trial (0.41 for type A blood).
• \( (1-p)^{k-r} \) accounts for the probability of the remaining trials being unsuccessful.

By substituting different values of \( k \), you can compute probabilities for varying numbers of donors needed, showing a comprehensive picture of how many donors are likely to be needed.

Expected Value

The Expected Value in probability gives us a "long-run" average for a random variable. In the context of the Negative Binomial Distribution, it helps predict the average number of trials needed.

For our blood donor scenario, the expected value \( \mu \) tells us how many donors, on average, the clinic can expect to need. The formula for the expected value in a Negative Binomial Distribution is:
\[ \mu = \frac{r}{p} \]
Where:

• \( r \) is the number of completed successes needed, which is obtaining 3 pints.
• \( p \) is the probability of success during each donor trial, here it’s 0.41 since type A is found in 41% of the population.

Plugging in these numbers, the expected value \( \mu \) is approximately 7.3171. This means the clinic can expect to see about 7.3 donors to collect the desired 3 pints of type A blood. Of course, you can’t have a fraction of a donor, but it illustrates that sometimes you will need more, or sometimes less than exactly 7 donors!

Standard Deviation

Standard Deviation provides insight into the variability or dispersion for the random variable around its mean, illuminating how spread out the donors' count might be.

In this context, it tells us how much the actual number of donors it takes to obtain 3 pints of blood might vary from the average. For the Negative Binomial Distribution, the standard deviation \( \sigma \) is calculated using this formula:
\[ \sigma = \sqrt{\frac{r(1-p)}{p^2}} \]
Where:

• \( r \) is again the total successes we want, which is 3 pints of blood.
• \( p \) is the probability of any single donor having type A blood.
• \( 1-p \) is the probability of a donor not having type A blood.

By substituting into the formula, we get a standard deviation \( \sigma \) of about 2.7077. This indicates that while the average number of donors needed is around 7, there might typically be a variance of about 2.7 donors in either direction to achieve the goal of 3 pints of blood. So, it largely prepares the clinic for fluctuations in donor numbers.