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Suppose that $s=\{a, b\}$ where $a$ is the event that Tom takes a pill on a single day and $b$ is the event that Tom doesn't take a pill on a single day.

Let $$ X(x)=\left\{\begin{array}{ll} 1 & x=a \\ 0 & x = b \end{array}\right. $$

and $$ \begin{array}{r} P(X=1) =0.9 \\ P(X=0) =0.1 \end{array} $$

If we track this random variable for a year, can we say that the distribution approaches a normal distribution even though it was initially highly skewed to to the left. Would this only hold if the events were independent?

What is there was some weak dependence of the events realistically speaking. Can we still use CLT? I guess since $n=1$, the average of a number is just the number. But would we need more than 365 repetitions to use CLT (assuming we can even use it)?

Cltprobguy
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    The CLT has to do with sums of RV, you do not have any sum here, only X, which is binary, so this will never approach any kind of normal distribution. – user2974951 Dec 15 '21 at 08:08
  • I am sampling from a distribution and each sample is of size 1. Are you saying that I would have to sample say, 2 times each day take the average and repeat this for a year so that the sample means would be normally district? – Cltprobguy Dec 15 '21 at 08:24
  • I don't know, you will have to define what you are interested in summing. You could for ex. sum all the daily outcomes for each week, and this should approach a normal distribution (the number of pills taken in a week). – user2974951 Dec 15 '21 at 08:35
  • So today I could get 100 samples where n=2, (0,1 --> 1, (1,0) -->1, (1,1) -->2 etc.. tomorrow another 100 samples where n = 2, (0,0)--> 0, (1,1) --> 2 etc. so I am summing the daily outcomes for n =2 for 100 samples each day. – Cltprobguy Dec 15 '21 at 10:10
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    [Do you have the same common misconception about the central limit theorem that I once had?](https://stats.stackexchange.com/questions/473455/debunking-wrong-clt-statement) – Dave Dec 15 '21 at 10:56

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the distribution does not approach a normal distribution (it predetermined as a bernoulli distribution).

but the normalized sum does approach a normal distribution:

lets define: z = (x1+x2+...+x365)/365

so intuitively, you would expect E(z) ~= 0.9

but this this new z could also be 0.899 or 0.902 for example. so there is some variance for this new term. you can show that var(z) = sigma/sqrt(365), where sigma is the variance of your original distribution (for bernuli dist, sigma is p(1-p)).

I would suggest seeing some youtube vids on this subject (with visualizations)

good luck!

Jonathan
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  • So today I could get 100 samples where n=2, (0,1 --> 1, (1,0) -->1, (1,1) -->2 etc.. tomorrow another 100 samples where n = 2, (0,0)--> 0, (1,1) --> 2 etc. so I am summing the daily outcomes for n =2 for 100 samples each day. – Cltprobguy Dec 15 '21 at 10:11