Tagged: CENTRAL LIMIT THEOREM
 This topic has 2 replies, 1 voice, and was last updated 2 months ago by Sarthak Goel.
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July 30, 2020 at 1:37 pm #26677Sarthak GoelParticipant
Central Limit Theorem states that the sample mean follows a normal distribution with mean equals to the population mean and the variance equals population variance divided by the number of samples.
Now the question is have you just cramped it or have you ever tried to see whether this is actually true or not.
July 30, 2020 at 1:51 pm #26680Sarthak GoelParticipantSo, here is a little exercise that would solve the problems of people scratching their minds for the proof of CLT.
The sum on rolling of 2 dice
Step 1: Take the sum on rolling of 2 dice 10 times. This is your first sample containing 10 values.
This can be done using Excel as follows:
=RANDBETWEEN(1,6)
This will give you a random number between 1 and 6 and the same function can be put in another column to get the value on the second dice. Drag it to 10 rows so that sum can be calculated for rolling of 2 dice 10 times.
Copy it and paste special it as values, this is because the randbetween function refreshes itself after every operation in the workbook.
Take the sum of the values received in 2 columns. Repeat it for all 10 rows.
Now, you have 10 values reflecting the sum on rolling of 2 dice.
Step 2: Take the mean of these 10 values by using the AVERAGE function.
This would be your first sample mean.
Step 3: Repeat the above process at least 30 times, this would be your n, i.e., the number of samples.
Now, you will be having 30 sample mean values.
Step 4: Take the mean of these 30 values using AVERAGE function and Variance by VAR.S function.
Step 5: Calculate the mean and variance of the sum on rolling of dice 2 times using probabilities.
The answer of Step 5 would be same for everyone, these are the population parameters and the values would be:
Population mean= 7
Population variance= 5.83
Step 6: Compare the mean in Step 4 with that of Step 5 and variance in step 4 with variance in step 5 divided by n, i.e., 30 in this case.
BINGOOOOO!!!!!!
It is almost equal. You will see that the difference between the values in 2 steps would decrease as you would increase the number of samples, thus smoothening the normal curve.
Thus, we have proved that as the number of samples increases and becomes large, the distribution of the sample mean would follow a normal distribution with mean that equals the population mean and variance being equal to population variance divided by the number of samples.
MUST TRY IT!!!!
IT IS WORTH INVESTING YOUR TIME!!! 
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