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.
(a) The formula for P(n) in the context of this application, which represents the probability that the fourth successful crop occurs in year n, can be calculated using the negative binomial distribution formula:
P(n) = ( k-1 C n-1 ) * p^k * q^(n-k)
Where:
- ( k-1 C n-1 ) represents the combination of k-1 choose n-1.
- p is the probability of success on each trial (0.65 in this case).
- q is the probability of failure on each trial (0.35 in this case).
- k is the fixed number of successful crops needed to repay the loan (4 in this case).
(b) To compute P(n=4), P(n=5), P(n=6), and P(n=7), we substitute the values into the formula and solve for each case:
P(n=4) = (3 C 3) * (0.65^4) * (0.35^(4-4)) = 0.65^4 ≈ 0.179
P(n=5) = (4 C 4) * (0.65^4) * (0.35^(5-4)) = 0.65^4 * 0.35 ≈ 0.073
P(n=6) = (5 C 5) * (0.65^4) * (0.35^(6-4)) = 0.65^4 * 0.35^2 ≈ 0.030
P(n=7) = (6 C 6) * (0.65^4) * (0.35^(7-4)) = 0.65^4 * 0.35^3 ≈ 0.013
(c) To find the probability that the Wagners can repay the loan within 4 to 7 years, we sum the probabilities P(n=4) to P(n=7):
P(4 to 7 years) = P(n=4) + P(n=5) + P(n=6) + P(n=7)
= 0.179 + 0.073 + 0.030 + 0.013 ≈ 0.295
So, there is approximately a 29.5% chance that the Wagners can repay the loan within 4 to 7 years.
(d) To find the probability that the Wagners will need to farm for 8 or more years before repaying the loan, we can calculate P(n is greater than or equal to 8) as follows:
P(n ≥ 8) = 1 - [P(n=4) + P(n=5) + P(n=6) + P(n=7)]
= 1 - 0.295 ≈ 0.705
So, there is approximately a 70.5% chance that the Wagners will need to farm for 8 or more years before repaying the loan.
(e) The expected value (μ) and standard deviation (σ) of the random variable n can be calculated using the formulas:
μ = k/p,
σ = sqrt((k*q) / (p^2))
Using the given values, we have:
μ = 4 / 0.65 ≈ 6.15
σ = sqrt((4 * 0.35) / (0.65^2)) ≈ 1.8
In this context, the expected value (6.15) represents the average number of years it would take for the Wagners to repay the loan, while the standard deviation (1.8) measures the variability or spread of the number of years around the expected value.