The number of combines sold each year by a dealership that specialises in agricultural equipment is a Poisson random variable with an average of 4.
What is the probability that the dealership will sell:
3.1 Less than five combines in a given year.
3.2 More than three combines in a period of six months.
To solve these problems, we can use the Poisson probability formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
P(X = k) is the probability of selling k combines in a given time period
e is the base of the natural logarithm (approximately 2.71828)
λ is the average number of combines sold in a given time period
k is the number of combines we want to find the probability for
3.1) Probability of selling less than five combines in a given year:
λ = 4
k = 0, 1, 2, 3, 4
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
P(X = 0) = (e^(-4) * 4^0) / 0! = 0.0183
P(X = 1) = (e^(-4) * 4^1) / 1! = 0.0733
P(X = 2) = (e^(-4) * 4^2) / 2! = 0.1465
P(X = 3) = (e^(-4) * 4^3) / 3! = 0.1954
P(X = 4) = (e^(-4) * 4^4) / 4! = 0.1954
P(X < 5) = 0.6289 or 62.9%
3.2) Probability of selling more than three combines in a period of six months:
Since the average is given annually, we need to adjust it for six months:
λ = 4 * (6/12) = 2
k = 4, 5, 6, ...
P(X > 3) = 1 - P(X ≤ 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
P(X = 0) = (e^(-2) * 2^0) / 0! = 0.1353
P(X = 1) = (e^(-2) * 2^1) / 1! = 0.2707
P(X = 2) = (e^(-2) * 2^2) / 2! = 0.2707
P(X = 3) = (e^(-2) * 2^3) / 3! = 0.1805
P(X > 3) = 1 - (0.1353 + 0.2707 + 0.2707 + 0.1805) = 0.1428 or 14.3%
Therefore, the probability that the dealership will sell:
3.1) Less than five combines in a given year is 62.9%
3.2) More than three combines in a period of six months is 14.3%