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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 $E(x)=9/4$

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

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  • $\begingroup$ Hint: what is the definition of expected value? $\endgroup$ – Sean Roberson Apr 22 '18 at 0:50
  • $\begingroup$ @SeanRoberson $E(x)=\sum xp(x)$, what would be p(x) in this case? $\endgroup$ – user441848 Apr 22 '18 at 0:56
  • $\begingroup$ 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. $\endgroup$ – Rivaldo Apr 22 '18 at 1:00
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    $\begingroup$ @Rivaldo oh ok so $E(x)=1/4+1/2+3/2=2+1/4=9/4$ $\endgroup$ – user441848 Apr 22 '18 at 1:08
  • $\begingroup$ Personally i think this question belongs to Cross-Validated $\endgroup$ – Victor S. Apr 22 '18 at 1:09
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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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    $\begingroup$ Thank you so much Mee Seong Im, your answer is very clear. $\endgroup$ – user441848 Apr 22 '18 at 3:00

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