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 $E(x)=9/4$
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.