Zachary C.

# How can I estimate the probability of a random variable from one population being greater than all other random variables from unique populations?

Lets assume I have samples from 4 unique populations. Let's also assume I have a mean and standard deviation from each of these populations, they are normally distributed and completely independent of one another.

How can I estimate the probability that a sample of one of the populations will be greater than a sample from each of the other 3 populations?

For a example, if I have 3 types of fish (the populations) in my pond, such as bass, catfish, karp, and koi, and i'm measuring the lengths (the variables) of the fish, how do can I estimate the probability that the length of a bass I catch will be greater than the length of all the other types of fish? I think I understand how to compare 2 individual populations but can't seem to figure out how to estimate probability relative to all populations. As opposed to the probability of the bass to a catfish, and then a bass to a karp, etc., I'd like to know if its possible to reasonably estimate the probability of the length of the bass being greater that the lengths of all other populations.

For example if the fish have means and standard deviations of:

Bass (40, 10)

Catfish (36, 8)

Karp (45, 12)

Koi (35, 15)

I would think that finding the probability of the Bass being greater than all other fish would be the product of the individual probabilities P(Y>X1) multiplied by P(Y>Xn). However, if I calculate the probability the a fish will be greater than all others for each type of fish (probability a bass is the greatest length,probability catfish is the greatest length, etc) wouldn't the sum of these probabilities be equal to 1? Surely 1 fish will be greater than all others.

## 3 Answers By Expert Tutors

By: Tutor
4.7 (31)

Math and computer tutor/teacher

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.