Variance of 36 standard dice rolls

[Haatajajunk]

So i need to find the variance of a number of dice rolls. I know that i use the following formula:


N
∑ ( X i - µ)2
i = 1
σ2 = ---------------
N

but i don't know how to use it. I feel really stupid asking this but i am really lost right now. Thanks for any help in advance.

 

[Claude Bile]

I think N stands for the number of samples (i.e. 36), mu is the mean (which you can calculate easily enough) and X is the value for the ith sample.

What you are essentially doing is finding the average deviation from the mean.

You would do well to double check my answer though.

Claude.

 

[M98Ranger]

Don't worry, it's completely normal to feel lost when encountering a new formula or concept. Let's break down the formula and see how we can apply it to finding the variance of 36 standard dice rolls.

First, let's define the variables in the formula. N represents the number of observations, in this case the number of dice rolls which is 36. Σ (sigma) is the symbol for summation, meaning we will be adding up a series of numbers. X represents the individual dice rolls, and µ (mu) represents the mean or average of the dice rolls.

Now, let's plug in the numbers into the formula. We have N = 36, so our formula becomes:

σ2 = (1/36) * ∑ (Xi - µ)2

Next, we need to find the mean of the dice rolls. Since we are dealing with standard dice rolls, we know that the mean is 3.5 (calculated by adding all the possible outcomes of a dice roll and dividing by 6). So, our formula becomes:

σ2 = (1/36) * ∑ (Xi - 3.5)2

Now, we need to plug in the values for each dice roll. Let's say our 36 dice rolls are: 4, 2, 6, 1, 5, 3, 6, 2, 4, 1, 5, 3, 6, 2, 4, 1, 5, 3, 6, 2, 4, 1, 5, 3, 6, 2, 4, 1, 5, 3, 6, 2, 4, 1, 5, 3, 6. Our formula now becomes:

σ2 = (1/36) * [(4-3.5)2 + (2-3.5)2 + (6-3.5)2 + (1-3.5)2 + (5-3.5)2 + (3-3.5)2 + (6-3.5)2 + (2-3.5)2 + (4-3.5)2 + (1-3.5)2 + (5-3.5)2 + (3-3.5)2 + (6-