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A fair four-sided die with four equilateral triangle-shaped faces is tossed 200 times. Each of the die's four faces shows a different number from 1 to 4.

(a) Find the expected value of the sample mean of the values obtained in these 200 tosses.

(b) Find the standard deviation of the number obtained in 1 toss.


I think the wording of (a) is a bit confusing to me. It feels like this is a binomial distribution, with $n = 200, p = .25$. Thus $E(X) = np = 200\cdot.25 = 50$. However in this context, I am not sure what the sample mean refers to. Is the sample mean also $200$?

For (b), if my reasoning was correct for (a), then would I simply find $Var(X) = np(1-p)$ when $n = 1$? so $Var(X) = .25(1-.25) = .1875$, then $\sigma = \sqrt{.1875}$?


Edit: Update on work

(a) So I believe the population variables are $X_1 = 1, X_2 = 2, X_3 = 3, X_4 = 4$. Thus there are only $n=4$ members in the population of the four-sided dice and we are sampling this population 200 times. Then the coefficients should also be $P_1 = .25, P_2 = .25, P_3 = .25, P_4 = .25$.

Then \begin{align}E(\bar{X}) &= E[(P_1X_1 + P_2X_2 + P_3X_3 + P_4X_4)] \\ &= P_1E[X_1] + P_2E[X_2] + P_3E[X_3] + P_4E[X_4] \\ &=.25[1] + .25[2] + .25[3] + .25[4] \\ &= 2.5 \end{align}

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    $\begingroup$ For a), $E[\bar{X}] = E[(X_1 +...+X_n)/n]$ in general. Can you work out the expectation in your case? More fundamentally, do you understand what a sampling distribution is? (And please use the self-study tag.) $\endgroup$ – Dave Nov 9 '19 at 3:37
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    $\begingroup$ What are the potential outcomes in each toss? Can this be a binomial distribution? $\endgroup$ – Stephan Kolassa Nov 9 '19 at 6:07
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    $\begingroup$ I'd avoid paying a lot of attention to your feelings when it comes to stats/probability -- It has a tendency not to act in accordance with most people's intuition. Instead rely on the definition of variance and basic facts about variances. $\endgroup$ – Glen_b -Reinstate Monica Nov 9 '19 at 6:43
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I think you should follow your textbook. For a fair dice, the probability of each number is: $P(1)=P(2)=P(3)=P(4)=0.25$. The expected value should be $$E(X)=1\times0.25+2\times0.25+3\times0.25+4\times0.25 = 2.5$$.

Then you use the formula provided by Dave in the comment. You should get the same results of 2.5.

Similarly, $$Var(X) = E((X-\mu)^2) = \frac{(1-2.5)^2+(2-2.5)^2+(3-2.5)^2+(4-2.5)^2}{4}=1.25$$ for one tose.

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  • $\begingroup$ oh. I just updated my post, okay I was slightly off with $E(X)$ and I think completely off for my thinking of (b), thank you $\endgroup$ – Evan Kim Nov 9 '19 at 17:52
  • $\begingroup$ @EvanKim Getting back to my comment, you know the expectation of each $X_i$ (binomial expected value). Use the linearity of the expectation operator: $E[aX + bY] = aE[X]+bE[Y]$. (Hint: the “divide by n” is the same as “multiply by 1/n”.) $\endgroup$ – Dave Nov 9 '19 at 18:00
  • $\begingroup$ Ya I think I just got confused and did the opposite of what I was supposed to do. What I wrote down on paper was what Bill did for $E(X)$, just somehow I started thinking crazy and wrote something weird on here $\endgroup$ – Evan Kim Nov 9 '19 at 18:02
  • $\begingroup$ Just one more comment, I think that $VAR(X)$ should technically be $\frac{1}{n} \sum_{i=1}^{n=4}E[(X_i - \mu)^2]$ $\endgroup$ – Evan Kim Nov 9 '19 at 18:09
  • $\begingroup$ @EvanKim If you’re adding up four Xs in your expectation of $\bar{X}$ instead of 200, you’re still making a mistake. You’ll happen to get the right answer either way, but you are doing it wrong. $\endgroup$ – Dave Nov 9 '19 at 18:09

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