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From a population $A$ which follows a normal distribution with mean $μ= 100$ and standard deviation $σ = 5$ we take a sample $n(A) = 164$. From a population B which follows a normal distribution with mean $μ= 102$ and standard deviation $σ = 10$ we take a sample $n(B)= 25$ Calculate the probability that the mean of the sample from population $A$ is larger from the mean of the sample from population $B$?

Because the sample is less than $30$ I guess normal rules of Central Limit Theory do not apply. This plus the fact that I can only find solved examples of problems that calculate greater/less than about one population have me at a loss. Can anyone help me solve this? I am severely stuck. Thank you for any help.

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Slightly bigger hint: The mean of a sample of normally distributed values is again normally distributed, now with $\mu_M = \mu$ and $\sigma_M = \frac{\sigma}{\sqrt{n}}$. If we do this for both we now have two new normal distributions, let's call them $X_A$ and $X_B$. We are looking for $P(X_A>X_B)$ which is the same as $P(X_A-X_B>0)=P(X_B-X_A<0)$.

$X_A\sim N\left(100,\left(\frac{5}{\sqrt{164}}\right)^2\right), X_B\sim N\left(102,\left(\frac{10}{\sqrt{25}}\right)^2\right)$

$X_B-X_A$ is again normally distributed. $X_B-X_A = X_C \sim N\left(2,\left(\frac{5}{\sqrt{164}}\right)^2+\left(\frac{10}{\sqrt{25}}\right)^2\right)$

We are looking for $P(X_A>X_B) = P(X_C<0)$

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  • $\begingroup$ Does that mean that this is normally distributed with mean μ and variance σ^2/n. With probability distribution of the sample mean (from population A) being : A(μ,σ^2/n) ? Edit: Wrote that before seeing your edit. Thank you very much. $\endgroup$
    – apot
    Commented Dec 13, 2015 at 20:58
  • $\begingroup$ Extended it a little bit further again! No problem :) $\endgroup$ Commented Dec 13, 2015 at 21:04

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