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While doing the dice rolls, we divide "by the standard error".
|================ | 22%
| To show the CLT in action consider this figure from the slides. It presents 3
| histograms of 1000 averages of dice rolls with sample sizes of 10, 20 and 30
| respectively. Each average of n dice rolls (n=10,20,30) has been normalized by
| subtracting off the mean (3.5) then dividing by the standard error, sqrt(2.92/n).
| The normalization has made each histogram look like a standard normal, i.e., mean
| 0 and standard deviation 1.
But then when doing coin flips, we divide "by the std deviation".
|================== | 25%
| Recall that if the probability of a head (call it 1) is p, then the probability of
| a tail (0) is 1-p. The expected value then is p and the variance is p-p^2 or
| p(1-p). Suppose we do n coin flips and let p' represent the average of these n
| flips. We normalize p' by subtracting the mean p and dividing by the std deviation
| sqrt(p(1-p)/n). Let's do this for 1000 trials and plot the resulting histogram.
The text was updated successfully, but these errors were encountered:
Course: Statistical Inference
Lesson: Asymptotics
While doing the dice rolls, we divide "by the standard error".
But then when doing coin flips, we divide "by the std deviation".
The text was updated successfully, but these errors were encountered: