# How to find mean and standard deviation? [closed]

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 $E(x)=9/4$

## closed as off-topic by LinAlg, Saad, JonMark Perry, Claude Leibovici, ShaileshApr 22 '18 at 11:27

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• Hint: what is the definition of expected value? – Sean Roberson Apr 22 '18 at 0:50
• @SeanRoberson $E(x)=\sum xp(x)$, what would be p(x) in this case? – user441848 Apr 22 '18 at 0:56
• p(x=1)=1/4... yes like you said above since its a discrete random variable. So X=1 X=2 X=3 are the possible values. – Rivaldo Apr 22 '18 at 1:00
• @Rivaldo oh ok so $E(x)=1/4+1/2+3/2=2+1/4=9/4$ – user441848 Apr 22 '18 at 1:08
• Personally i think this question belongs to Cross-Validated – Victor S. Apr 22 '18 at 1:09

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.