<s>[INST] How to find mean and standard deviation? An urn contains a large number of cards of which 1/4 have the number 1,1/4 have the number 2 and 1/2 have the number 3.
a) Let $X$ be the number of the card when a card is taken from the urn. Find the mean and standard deviation of $X$.
b) Let $X$ be the sample mean when card samples are taken, compute $\mu_{\overline x}$ and $\sigma_{\overline x}$.
Any hint please I don't know what to do

Edit: By definition $E(x)=\sum xp(x),$ thus $E(x)=9/4$
 [/INST] Your probability mass function (pmf) $p$ is 
$$
p(x)=P(X=x)=
\begin{cases}
\frac{1}{4} &\mbox{ if } x=1, \\ 
\frac{1}{4} &\mbox{ if } x=2, \\ 
\frac{1}{2} &\mbox{ if } x=3. \\
\end{cases}
$$
a.) The expected value (mean) of $X$ is 
$$
\mu=\mu_X = E[X] = \sum_{x=1}^{3} xp(x) = 1\left(\frac{1}{4}\right)+ 2\left(\frac{1}{4}\right)+ 3\left(\frac{1}{2}\right) = \frac{9}{4}
$$
and the variance of $X$ is 
$$
V(X) = \sum_{x=1}^{3}(x-\mu)^2 p(x) 
= \left(1-\frac{9}{4}\right)^2 \frac{1}{4} + \left(2-\frac{9}{4}\right)^2 \frac{1}{4} + \left(3-\frac{9}{4}\right)^2 \frac{1}{2} = \frac{11}{16}.
$$
So the standard deviation of $X$ is 
$$
\sigma = \sigma_X = \sqrt{V(X)} = \sqrt{\frac{11}{16}}=\frac{\sqrt{11}}{4}. 
$$
b.) We have 
$$
\mu_{\overline X} = \mu_X = \frac{9}{4} 
$$
while 
$$
\sigma_{\overline X} = \frac{\sigma_X}{\sqrt{n}} = \frac{\sqrt{11}}{4\sqrt{n}}, 
$$
where $n$ is the sample size. 
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