Weighted means, medians and probability

Give each observation one vote

Ten students have a mean score of 28. Another thirty students have a mean of 20. Before doing any arithmetic, mark 20 and 28 on a number line. Which end should the combined mean lie nearer? Three quarters of the people come from the second group. Giving the two group means an equal say would make one person in the small group count as much as three people in the large group.

A weighted mean (media ponderata) keeps track of how many observations each number represents. Multiply each mean by its group size to recover the sum of that group's scores. Combining the groups combines those sums. Divide the resulting total by the number of people who contributed to it. You can do this even when nobody's individual score is known. The information lost in taking a mean does matter for some questions, but the total survives as long as you know the count.

Here there is also a short mental route. Give everyone 20 points provisionally. The ten students in the first group contribute an extra eight points each on average. Sharing their eighty extra points across forty people adds two points per person. The combined mean is 22, one quarter of the way from 20 to 28. This difference method is especially convenient when all the numbers in a table share a large common starting value.

Locate the middle observation

The median (mediana) uses position. Order the observations from smallest to largest. In a list of nine entries, four lie on either side of the fifth position, so read the fifth entry. For ten entries the centre lies halfway between positions five and six. Average the values at those positions. If they coincide, their average is that same value.

A frequency table stores a sorted list in compressed form. Suppose two seedlings have three leaves, four have five leaves and three have eight leaves. The first two positions carry the value three. Positions three through six carry five, and the remaining three carry eight. Position five therefore gives a median of five leaves. The group means in the student problem tell us too little to reconstruct this sort of ranking. Different lists of individual marks can give exactly the same total.

Count outcomes once

For a finite experiment with equally likely outcomes, divide the number that satisfy the event by the number possible. On a fair six-sided die the event “two or odd” includes four faces. The face showing two cannot also show an odd number, so the two lists can be added directly. Events which cannot occur together are called mutually exclusive (incompatibili).

Change the first event to “prime”. The primes on this die are two, three and five. Two of them are also odd. Adding the sizes of the prime and odd lists would count three twice and five twice. Remove one copy of each repeated outcome. In symbols the inclusive “or” has probability $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. This identity also holds when outcomes have unequal chances. In that case use their probabilities rather than counting them as though each had the same weight.

Decide where a product belongs

Events are independent (indipendenti) when knowing that one occurred leaves the probability of the other unchanged. For independent events the chance of both is the product of their separate probabilities. For instance, spin a fair two-colour spinner and independently roll the die. Each spinner result can occur alongside every die result. The chance of red and an odd number is half of one half.

To find the chance of at least one success, you can count the single case in which both events fail and subtract its probability from one. With independent trials you may multiply their failure probabilities. With dependent events you need the chance of the second failure given the first, or an explicit list of possible outcomes. As a final check, a union must be at least as probable as either event alone.