@ГригорийПерельман If you elaborate a bit more, maybe someone can help you. Not everyone here is into statistics.

I have a given normal distribution, from it I need to get a sampling distribution of means with sample size 30.

Except I have no idea how to do that.

@ГригорийПерельман OK, so you have a "sample" of a normally distributed data and now you want to take a random sample from it and calculate the means?

No, I have a NormalDistribution (NormalDistribution[17, 0.8]) and I need to get a distribution of all possible samples of size 30.

@ГригорийПерельман When I read this correctly here, then you can just get what you want by

NormalDistribution[mu, sigma^2/30]

where mu and sigma is 17 and 0.8 resp. Not?

Maybe. I'm not a very smart person, I was hoping there's a builtin way.

@ГригорийПерельман Give me one sec..

Thanks, by the way.

@ГригорийПерельман OK, let us create some data. Let's say we have your normal distribution and we take 10000 samples of size 30 and calculate their mean:

distr = NormalDistribution[17, 0.8];
means = Table[Mean@RandomVariate[distr, 30], 10000];

Thanks!

@ГригорийПерельман Now, we can look at the histogram of all those means:

Yep, I can do that myself. Thanks for telling me about the RandomVariate function!

My Google-Fu has been really off today.

Assuming now, that the means are again normally distributed, we can stubbornly try to fit a normal distribution on it.

In[210]:= FindDistributionParameters[means, NormalDistribution[mu, sigma]]

Out[210]= {mu -> 16.9998, sigma -> 0.144976}

You see we get a mu of about 17 and a sigma of 0.145

On a different website they say that the distribution of the means is a normal distribution with mu2=my and sigma2=sigma/sqrt(n) where mu and sigma are 17 and 0.8. So let's test this:

I only calculate sigma2:

In[212]:= 0.8/Sqrt[30]

Out[212]= 0.146059

@ГригорийПерельман So your "built-in" function you want is nothing more than

normalDistributionOfMeans[mu_, sigma_, n_] := NormalDistribution[mu, sigma/Sqrt[n]]

Ok! I have to go to class now, and I think you just saved my life there.

@ГригорийПерельман Good luck.

is there some sort of built-in NLP-ish loosey goosey string date parser? something that might understand "last night" perhaps. i tried Interpreter["Date"]["last night"] and it didn't get it - i think its looking for more of a date literal. i thought i had seen something in the examples but having no luck finding it again. strtotime in php is a simple analog of what im seeking (though perhaps too limited for this example)

@ThomasLackner What about

SemanticInterpretation["Last night"]

@halirutan color me impressed. cant believe i missed it. thanks

This is a duplicate but I cannot find the original.

Could someone on Windows post the value of $Version for v11.1.0 here?

@user6014 Do you remember on what date you downloaded it?

@Szabolcs 18th

Thanks

#daily-fun

@halirutan @Григорий Almost: Mathematica's NormalDistribution function expects the second parameter to be the standard deviation rather than the variance. So NormalDistribution[mu, sigma/Sqrt[30]] will get one the sampling distribution of the sample mean from a sample of size 30 from a NormalDistribution[mu, sigma]`.

@halirutan Sorry. I see that you use the correct form in the subsequent messages.

@JimBaldwin Yes, I saw my mistake later. Nevertheless, thanks for pointing it out.