statistics Q

stan56

tryin to do a question here but cant really understand it
first of all how do you calculate variance?
also in the question i am doing there is a table of payout and probability and it asks for the mean and variance of the payout, is this just asking for the average payout and how do i calculate this?
Any help appreciated

2Scoops

Variance is just a way of expressing the distribution of scores about the mean (the "average"). It is calculated by finding the distance of the scores from the mean, squaring them individually, adding up all those numbers and then getting the mean of that.

MathsManiac

It's also worth mentioning that the variance is the same as the standard deviation squared. If you remember doing calculations of standard deviation at school, then just do the exact same thing for variance, except that you don't the very last step of getting the square root.

Calculating the mean and variance of a probability distribution is procedurally exactly the same as getting the mean and variance of a data set in a frequency table. It's just that you have probabilities instead of frequencies. (Except the probabilities always sum to 1, so you don't need to bother dividing by their sum.)

The mean of a probabilty distribution is usually referred to as its expected value. In the case of a "payout" as you seem to have, it represents the average payout per game in the long run, as you guessed.

e.g.: I roll a die. If it comes up 1,2, or 3, I win nothing, if it comes up 4 or 5, I win €1, and if it comes up 6, I win €2. I could represent this as:

outcome (payout).. €0......€1.....€2
probability....... 1/2.....1/3....1/6

The mean is 0*(1/2) + 1*(1/3) + 2*(1/6) = 2/3
(meaning that on average, I'd win 2/3 of a euro per game if I play for a long time, so it's a good bet if it costs me 50 cent to play, and a bad bet if it costs a euro to play.)

The variance is (0 - 2/3)^2*(1/2) + (1 - 2/3)^2*(1/3) + (2 - 2/3)^2*(1/6) = 5/9.

It's a bit harder to understand what this represents in the case of a probabilty distribution than in the case of a data set, but it's a measure of the expected variation in the outcome of the random process.

stan56

thanks for the replys guys i get the idea of it now