In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that the number of soldiers killed by horse kicks each year in each corps in the Prussian cavalry followed a Poisson distribution with a mean of 0.63.
(a) What is the probability of more than one death in a corps in a year?
(b) What is the probability of no deaths in a corps over 4 years?
0.2845
0.6984
(a) Ah, the law of small numbers. Sounds like it could use a little more excitement! To find the probability of more than one death in a corps in a year, we need to calculate the complement of the probability of zero or one death.
So, let's start by finding the probability of zero or one death:
P(X ≤ 1) = P(X = 0) + P(X = 1)
Since the data follows a Poisson distribution with a mean of 0.63 deaths per year, we can use the Poisson probability formula. The probability mass function is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
Where λ is the mean of the Poisson distribution and k is the number of deaths.
Plugging in the values, we get:
P(X ≤ 1) = P(X = 0) + P(X = 1)
= (e^(-0.63) * 0.63^0) / 0! + (e^(-0.63) * 0.63^1) / 1!
= e^(-0.63) + e^(-0.63) * 0.63
Now, to find the probability of more than one death, we simply subtract the above result from 1:
P(X > 1) = 1 - P(X ≤ 1)
(b) To find the probability of no deaths in a corps over 4 years, we first need to find the probability of no deaths in a single year, assuming the mean is 0.63.
Using the same Poisson probability formula, let's plug in the values:
P(X = 0) = (e^(-λ) * λ^0) / 0!
= e^(-0.63) * 0.63^0
Since the events are independent over the four years (the probability for each year doesn't change), we can simply raise the single-year probability to the power of four:
P(no deaths in 4 years) = P(X = 0)^4
And that should give you the probability of no deaths in a corps over 4 years. Remember, even though it's statistically interesting, it's always good to keep the real soldiers safe from horse kicks!
To solve these questions, we will use the Poisson distribution formula:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
- P(x; λ) is the probability of getting exactly x events
- λ is the mean number of events per interval (in our case, deaths per year)
- e is Euler's number, approximately 2.71828
- x is the number of events we're interested in finding the probability for
(a) To find the probability of more than one death in a corps in a year, we need to calculate the opposite probability, i.e., the probability of having zero or only one death.
For x = 0, P(0; 0.63) = (e^(-0.63) * 0.63^0) / 0! = e^(-0.63) ≈ 0.5314
For x = 1, P(1; 0.63) = (e^(-0.63) * 0.63^1) / 1! = 0.63 * e^(-0.63) ≈ 0.3347
So, the probability of having zero or only one death is approximately 0.5314 + 0.3347 = 0.8661.
To find the probability of more than one death, we subtract this probability from 1:
P(more than one death) = 1 - 0.8661 ≈ 0.1339
Therefore, the probability of more than one death in a corps in a year is approximately 0.1339.
(b) The probability of no deaths in a corps over 4 years is calculated by using the Poisson distribution for x = 0 and multiplying it for each year.
P(no deaths in 4 years) = P(0; 0.63) * P(0; 0.63) * P(0; 0.63) * P(0; 0.63) = e^(-0.63) * e^(-0.63) * e^(-0.63) * e^(-0.63) ≈ 0.2160
Therefore, the probability of no deaths in a corps over 4 years is approximately 0.2160.
To calculate the probability of more than one death in a corps in a year, we need to use the Poisson distribution formula. The probability mass function of the Poisson distribution is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- X corresponds to the number of deaths in a corps in a year.
- λ is the mean of the Poisson distribution, which is 0.63 in this case.
- k is the actual number of deaths.
To calculate the probability of more than one death, we need to sum up the probabilities for each value of k greater than one. Since the Poisson distribution is defined for discrete random variables, we can use the cumulative distribution function (CDF) to calculate this.
(a) The probability of more than one death in a corps in a year is:
P(X > 1) = 1 - P(X ≤ 1)
= 1 - [P(X = 0) + P(X = 1)]
To calculate P(X = 0), we substitute k = 0 into the Poisson distribution formula:
P(X = 0) = (e^(-0.63) * 0.63^0) / 0!
And to calculate P(X = 1), we substitute k = 1:
P(X = 1) = (e^(-0.63) * 0.63^1) / 1!
Once we have these probabilities, we can subtract them from 1 to find the probability of more than one death.
(b) To find the probability of no deaths in a corps over 4 years, we can use the fact that the number of deaths in each year follows a Poisson distribution and is independent from year to year. Therefore, we can use the formula for the product of independent events to calculate this probability.
The probability of no deaths in a corps over 4 years is:
P(no deaths in 4 years) = P(X = 0 in year 1) * P(X = 0 in year 2) * P(X = 0 in year 3) * P(X = 0 in year 4)
Each of these probabilities can be calculated using the Poisson distribution formula with λ = 0.63 for each year. Remember to use the formula for P(X = 0) when k = 0 in the Poisson distribution formula.
By following these steps, you can calculate the probabilities of more than one death in a corps in a year and no deaths in a corps over 4 years.
Poisson distribution (m = mean):
P(x) = e^(-m) m^x / x!
Values for (a):
x = 0, 1
m = 0.63
Substitute and calculate for each x. Add those calculations together for a total probability. Then subtract that value from 1. This will give you the probability of more than one death.
For (b):
To determine the number of deaths over 4 years, take (0.63)(4) = 2.52 (this is the mean). Now calculate P(0) using the above formula.
I hope this will help get you started.