When Central Limit Theorem breaks down












2












$begingroup$


Let say I have following numbers

4,3,5,6,5,3,4,2,5,4,3,6,5

I sample some of them, say, 5 of them, and calculate sum of 5 samples.
Then I repeat that over and over to get many sums, and I plot the values of sums in histogram, which will be Gaussian as Central Limit Theorem.



But when they are following numbers, I just replaced 4 with some big number,

4,3,5,6,5,3,10000000,2,5,4,3,6,5

Sampling sum of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians.



Is there any paper or research that mentioned this?
Thank you










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let say I have following numbers

    4,3,5,6,5,3,4,2,5,4,3,6,5

    I sample some of them, say, 5 of them, and calculate sum of 5 samples.
    Then I repeat that over and over to get many sums, and I plot the values of sums in histogram, which will be Gaussian as Central Limit Theorem.



    But when they are following numbers, I just replaced 4 with some big number,

    4,3,5,6,5,3,10000000,2,5,4,3,6,5

    Sampling sum of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians.



    Is there any paper or research that mentioned this?
    Thank you










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let say I have following numbers

      4,3,5,6,5,3,4,2,5,4,3,6,5

      I sample some of them, say, 5 of them, and calculate sum of 5 samples.
      Then I repeat that over and over to get many sums, and I plot the values of sums in histogram, which will be Gaussian as Central Limit Theorem.



      But when they are following numbers, I just replaced 4 with some big number,

      4,3,5,6,5,3,10000000,2,5,4,3,6,5

      Sampling sum of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians.



      Is there any paper or research that mentioned this?
      Thank you










      share|cite|improve this question









      $endgroup$




      Let say I have following numbers

      4,3,5,6,5,3,4,2,5,4,3,6,5

      I sample some of them, say, 5 of them, and calculate sum of 5 samples.
      Then I repeat that over and over to get many sums, and I plot the values of sums in histogram, which will be Gaussian as Central Limit Theorem.



      But when they are following numbers, I just replaced 4 with some big number,

      4,3,5,6,5,3,10000000,2,5,4,3,6,5

      Sampling sum of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians.



      Is there any paper or research that mentioned this?
      Thank you







      central-limit-theorem






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      share|cite|improve this question











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      asked 3 hours ago









      JimSDJimSD

      285




      285






















          2 Answers
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          1












          $begingroup$

          Let's recall, precisely, what the central limit theorem says.




          If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




          When we have a static list of numbers like



          4,3,5,6,5,3,10000000,2,5,4,3,6,5


          and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




          • Identically distributed is no problem: each number in the list is equally likely to be chosen.

          • Independent is more subtle, and depends on our sampling scheme. If we are sampling with replacement, then we violate independence. It is only when we sample without replacement that the central limit theorem is applicable.


          So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



          So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



          Three Normal Distributions



          Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



          (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
            $endgroup$
            – JimSD
            9 mins ago



















          2












          $begingroup$

          First of all, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. With the population of your first example, $30$ is in fact OK.



          pop <- c(4,3,5,6,5,3,4,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          But, with this second population, samples of, say, size $100$ are fine.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 100
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
            $endgroup$
            – guy
            3 hours ago












          • $begingroup$
            Perfect. Added. Tks.
            $endgroup$
            – Paulo C. Marques F.
            3 hours ago










          • $begingroup$
            Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
            $endgroup$
            – JimSD
            13 mins ago











          Your Answer





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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Let's recall, precisely, what the central limit theorem says.




          If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




          When we have a static list of numbers like



          4,3,5,6,5,3,10000000,2,5,4,3,6,5


          and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




          • Identically distributed is no problem: each number in the list is equally likely to be chosen.

          • Independent is more subtle, and depends on our sampling scheme. If we are sampling with replacement, then we violate independence. It is only when we sample without replacement that the central limit theorem is applicable.


          So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



          So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



          Three Normal Distributions



          Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



          (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
            $endgroup$
            – JimSD
            9 mins ago
















          1












          $begingroup$

          Let's recall, precisely, what the central limit theorem says.




          If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




          When we have a static list of numbers like



          4,3,5,6,5,3,10000000,2,5,4,3,6,5


          and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




          • Identically distributed is no problem: each number in the list is equally likely to be chosen.

          • Independent is more subtle, and depends on our sampling scheme. If we are sampling with replacement, then we violate independence. It is only when we sample without replacement that the central limit theorem is applicable.


          So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



          So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



          Three Normal Distributions



          Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



          (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
            $endgroup$
            – JimSD
            9 mins ago














          1












          1








          1





          $begingroup$

          Let's recall, precisely, what the central limit theorem says.




          If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




          When we have a static list of numbers like



          4,3,5,6,5,3,10000000,2,5,4,3,6,5


          and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




          • Identically distributed is no problem: each number in the list is equally likely to be chosen.

          • Independent is more subtle, and depends on our sampling scheme. If we are sampling with replacement, then we violate independence. It is only when we sample without replacement that the central limit theorem is applicable.


          So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



          So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



          Three Normal Distributions



          Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



          (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






          share|cite|improve this answer











          $endgroup$



          Let's recall, precisely, what the central limit theorem says.




          If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




          When we have a static list of numbers like



          4,3,5,6,5,3,10000000,2,5,4,3,6,5


          and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




          • Identically distributed is no problem: each number in the list is equally likely to be chosen.

          • Independent is more subtle, and depends on our sampling scheme. If we are sampling with replacement, then we violate independence. It is only when we sample without replacement that the central limit theorem is applicable.


          So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



          So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



          Three Normal Distributions



          Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



          (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          Matthew DruryMatthew Drury

          26k263105




          26k263105












          • $begingroup$
            Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
            $endgroup$
            – JimSD
            9 mins ago


















          • $begingroup$
            Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
            $endgroup$
            – JimSD
            9 mins ago
















          $begingroup$
          Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
          $endgroup$
          – JimSD
          9 mins ago




          $begingroup$
          Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
          $endgroup$
          – JimSD
          9 mins ago













          2












          $begingroup$

          First of all, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. With the population of your first example, $30$ is in fact OK.



          pop <- c(4,3,5,6,5,3,4,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          But, with this second population, samples of, say, size $100$ are fine.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 100
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
            $endgroup$
            – guy
            3 hours ago












          • $begingroup$
            Perfect. Added. Tks.
            $endgroup$
            – Paulo C. Marques F.
            3 hours ago










          • $begingroup$
            Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
            $endgroup$
            – JimSD
            13 mins ago
















          2












          $begingroup$

          First of all, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. With the population of your first example, $30$ is in fact OK.



          pop <- c(4,3,5,6,5,3,4,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          But, with this second population, samples of, say, size $100$ are fine.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 100
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
            $endgroup$
            – guy
            3 hours ago












          • $begingroup$
            Perfect. Added. Tks.
            $endgroup$
            – Paulo C. Marques F.
            3 hours ago










          • $begingroup$
            Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
            $endgroup$
            – JimSD
            13 mins ago














          2












          2








          2





          $begingroup$

          First of all, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. With the population of your first example, $30$ is in fact OK.



          pop <- c(4,3,5,6,5,3,4,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          But, with this second population, samples of, say, size $100$ are fine.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 100
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here






          share|cite|improve this answer











          $endgroup$



          First of all, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. With the population of your first example, $30$ is in fact OK.



          pop <- c(4,3,5,6,5,3,4,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 30
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here



          But, with this second population, samples of, say, size $100$ are fine.



          pop <- c(4,3,5,6,5,3,10000000,2,5,4,3,6,5)
          N <- 10^5
          n <- 100
          x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
          x_bar <- rowMeans(x)
          hist(x_bar, freq = FALSE, col = "cyan")
          f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
          curve(f, add = TRUE, lwd = 2, col = "red")


          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          Paulo C. Marques F.Paulo C. Marques F.

          17.2k35497




          17.2k35497








          • 1




            $begingroup$
            It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
            $endgroup$
            – guy
            3 hours ago












          • $begingroup$
            Perfect. Added. Tks.
            $endgroup$
            – Paulo C. Marques F.
            3 hours ago










          • $begingroup$
            Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
            $endgroup$
            – JimSD
            13 mins ago














          • 1




            $begingroup$
            It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
            $endgroup$
            – guy
            3 hours ago












          • $begingroup$
            Perfect. Added. Tks.
            $endgroup$
            – Paulo C. Marques F.
            3 hours ago










          • $begingroup$
            Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
            $endgroup$
            – JimSD
            13 mins ago








          1




          1




          $begingroup$
          It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
          $endgroup$
          – guy
          3 hours ago






          $begingroup$
          It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
          $endgroup$
          – guy
          3 hours ago














          $begingroup$
          Perfect. Added. Tks.
          $endgroup$
          – Paulo C. Marques F.
          3 hours ago




          $begingroup$
          Perfect. Added. Tks.
          $endgroup$
          – Paulo C. Marques F.
          3 hours ago












          $begingroup$
          Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
          $endgroup$
          – JimSD
          13 mins ago




          $begingroup$
          Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
          $endgroup$
          – JimSD
          13 mins ago


















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