Formula for Variance in statistics
So I've been watching this lecture on LinkedIn Learning, on statistics. On the subject of standard deviation for a discrete probability distribution table.
Please see below
To get the variance, the lecturer subtracts "the square of the mean" from the "WEIGHT TOTAL" (which is the sum of "# of Drinks, Squared" times the normalised weight or relative frequency).
I cannot, for the life of me, figure out how this works and how it yields the same result as the standard variance formula (sum of all data points' distance from the mean divided by "N-1").
Could someone please help me with this?
Thanks a bunch.
$\endgroup$ 21 Answer
$\begingroup$$$\sigma^2 \equiv \cfrac{\sum_{i=1}^n(x_i-\mu)^2}{n}$$
$$\sigma^2 = \cfrac{\sum_{i=1}^n(x_i^2-2\mu x_i+\mu^2)}{n}$$
$$\sigma^2 = \cfrac{\sum_{i=1}^nx_i^2}{n} - \cfrac{2\mu\sum_{i=1}^n x_i}{n}+\cfrac{\sum_{i=1}^n \mu^2}{n}$$
$$\sigma^2 = \cfrac{\sum_{i=1}^nx_i^2}{n} - 2\mu^2 + \mu^2 = \frac1n\sum_{i=1}^nx_i^2 - \mu^2 \\\square$$
If you are familiar with Expectation, this is equivalent to
$$Var(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2$$
For more details on the proof above, here is a Khan Academy video on the topic.
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