# How to compare means from a CLT exercise and a Normal distribution if i have $n=4896$ random samples

I have the central limit theorem exercise solved but the normal distribution not. i want to compare the means and now I am stucked.

the exercise is a discrete random variable modelling in a table with his $$E(x)$$, population mean and standart deviation. after that I have to run with $$4896$$ samples. all good the sum and average.

but i dont know how to compare this mean with the normal distribution. anyone knows?

• The question is confusing. What are you comparing? The mean is a number while the normal distribution is a two parameter function. Commented Feb 27, 2021 at 3:42
• i want to know how to calculate the mean of a normal distribution with n=4896 values and comparing with a central limit theorem exercise of a discrete random distribution with n=4896 values.. its like compare mean( normal ) == mean ( discrete random var ) the goal here is compare the theory with the aproximation of central limit theorem. Commented Feb 27, 2021 at 4:10
• It looks to me that you simply need to take the averages of the two data sets. I don't understand how the normal distribution comes into play. Do you have 4896 numbers from the normal distribution? Commented Feb 27, 2021 at 4:22

If you take the sample mean of each of your $$4896$$ samples, they should form a sampling distribution that is normally distributed with mean $$\mu$$ with $$\mu$$ being the population mean and standard deviation $$\sigma/\sqrt n$$ with $$\sigma$$ being the population standard deviation. That is, each of your samples has lots of observations in it. Once you take the average of the observations in each sample, and plot the resulting 4896 sample means, they should form an approximately normal distribution with aforementioned mean and standard deviation.