Expand Your Knowledge: Negative Binomial Distribution

Suppose you have binomial trials for which the probability of success on each trial is p and the probability of failure is q= 1-p. Let k be a fixed whole number greater than or equal to 1. Let n be the number of the trial on which the kth success occurs. This means that the first k-1 successes occur within the first n-1 trials, while the kth successes actually occurs on the nth trial. Now, if we are going to have k successes, we must have at least k trials.. So n = k, k+1, k+2, … and n is a random variable. The probability distribution for n is called the negative binomial distribution.

In eastern Colorado there are many dry land wheat farms. The success of a spring wheat crop is dependent on sufficient moisture in March and April. Assume that the probability of a successful wheat crop in the region is about 65%. So the probability of success in a single year is p=0.65 and the probability of failure is q = 0.35. The Wagner farm has taken out a loan and needs k = 4 successful crops to repay it. Let n be a random variable representing the year in which the fourth successful crop occurs (after the loan was made).

(a) Write out the formula for P(n) in the context of this application
(b) Compute P(n=4), P(n=5), P(n=6), and P(n=7)
(c) What is the probability that the Wagners can repay the loan within 4 to 7 years?
(d) What is the probability that the Wagners will need to farm for 8 or more years before they can repay the loan? Hint: Compute P(n is greater than or equal to 8)
(e) What are the expected value μ (mu) and standard deviation σ (sigma) of the random variable n? Interpret these values in the context of this application.

To solve this problem, we need to use the formula for the negative binomial distribution and apply it to the given scenario.

(a) The formula for P(n) in the context of this application is:

P(n) = C(n-1, k-1) * p^k * q^(n-k)

Where:
- C(n-1, k-1) represents the number of ways to arrange (k-1) successes within the first (n-1) trials.
- p is the probability of success in a single trial (0.65 in this case).
- q is the probability of failure in a single trial (1 - p = 0.35).
- k is the number of desired successes (4 in this case).
- n is the number of trials at which the kth success occurs (our random variable).

(b) To compute P(n=4), P(n=5), P(n=6), and P(n=7), substitute the values into the formula:

P(n=4) = C(4-1, 4-1) * 0.65^4 * 0.35^(4-4)
P(n=5) = C(5-1, 4-1) * 0.65^4 * 0.35^(5-4)
P(n=6) = C(6-1, 4-1) * 0.65^4 * 0.35^(6-4)
P(n=7) = C(7-1, 4-1) * 0.65^4 * 0.35^(7-4)

Calculate the combinations C(n-1, k-1) and raise p and q to the respective powers for each value of n to find the probabilities.

(c) To find the probability that the Wagners can repay the loan within 4 to 7 years, we need to sum the individual probabilities from n=4 to n=7:

P(repay within 4 to 7 years) = P(n=4) + P(n=5) + P(n=6) + P(n=7)

(d) To find the probability that the Wagners will need to farm for 8 or more years before repaying the loan, we need to calculate:

P(n is greater than or equal to 8) = 1 - (P(n=4) + P(n=5) + P(n=6) + P(n=7))

Take the complement of the sum of probabilities from part (c).

(e) To calculate the expected value μ (mean) of the random variable n, use the formula:

μ = k/p

In this case, μ = 4/0.65.

To calculate the standard deviation σ (standard deviation), use the formula:

σ = sqrt(k * q) / p

In this case, σ = sqrt(4 * 0.35) / 0.65.

Interpretation:
μ represents the average number of years it takes to achieve 4 successful crops, while σ represents the variation or spread around the mean. In the context of this application, μ tells us the average time it takes for the Wagners to repay their loan, and σ tells us how much the actual number of years may vary from the average.