We’re almost there! We have one small problem with our equation. Consider this very small dataset:

c(-5, 5)

The mean of this dataset is `0`

, so when we find the difference between each point and the mean we get `-5 - 0 = -5`

and `5 - 0 = 5`

.

When we take the average of `-5`

and `5`

to get the variance, we get `0`

:

`$\frac{-5 + 5}{2} = 0$`

Now think about what would happen if the dataset were `c(-200, 200)`

. We’d get the same result! That can’t possibly be right — the dataset with `200`

is much more spread out than the dataset with `5`

, so the variance should be much larger!

The problem here is with negative numbers. Because one of our data points was `5`

units below the mean and the other was `5`

units above the mean, they canceled each other out!

When calculating variance, if a data point was above or below the mean — all we care about is how far away it was. To get rid of those pesky negative numbers, we’ll square the difference between each data point and the mean.

Our equation for finding the difference between a data point and the mean now looks like this:

`$\text{difference} = (X - \mu)^2$`

### Instructions

**1.**

Square each of the values stored in the variables `difference_one`

through `difference_five`

. In R, to square a number, use the `^`

operator. The code below gives you an example of squaring the variable `a`

.

squared_num <- a ^ 2

# Take this course for free

By signing up for Codecademy, you agree to Codecademy's Terms of Service & Privacy Policy.