I am helping a friend with a study guide for a class, and one of the problems is asking about the theoretical mean and standard deviation. Two 8-sided dice with equal probabilities for 1, 2, 3, 4, 5, 6, 7, and 8, are rolled, and the sum of the two dice are recorded.
So I have a dataset that is the sum of the 8-sided dice rolled 10000 times. That is, I have a dataset of 10000 values between 2 and 16. This is what I did:
Mean: $\mu = \frac{\sum(x_i)}{N} = \frac{\sum_{x = 2}^{16}(\frac{10000}{15})(x)}{10000}$ = 9
Standard deviation: $\sigma = \sqrt{\frac{\sum(x_i-\mu)^2}{N}} = \sqrt{\frac{\sum_{x = 2}^{16}(\frac{10000}{15})(x-9)^2}{10000}} \approx$ 4.32
My friend's study guide has the numerical answers to all the questions, but they are not in order. I do not see 4.32 on this list of answers so I was wondering if I did it wrong.
Any help would be appreciated!
$\endgroup$2 Answers
$\begingroup$You have assumed that each sum from $2$ through $16$ is equally probable, so each one shows up $\frac 1{15}$ of the time. That is not true. There are $64$ different rolls and a sum of $9$ shows up $8$ times so has a chance of $\frac 18$ while $2$ and $16$ each only show up $\frac 1{64}$ of the time. Because of the symmetry your calculation of the mean came out, even though you calculated it incorrectly. The variance is smaller than you calculate.
What you should do is $\sum_{x=2}^{16}P(x)(x-9)^2$ where $P(x)$ is the probability of rolling a sum of $x$ This is $$\sqrt{\frac {1(2-9)^2+2(3-9)^2+3(4-9)^3+\ldots8(9-9)^2+7(10-9)^2+\ldots+1(16-9)^2}{64}}\approx 3.24$$
$\endgroup$ 1 $\begingroup$Refer to the table:$$\begin{array}{c|c|c|c} X&P(X) & XP(X)&X^2P(X^2)\\ \hline 2&1/64&2/64&4/64\\ 3&2/64&6/64&18/64\\ 4&3/12&6/64&48/64\\ \vdots\\ 8&7/64&56/64&448/64\\ 9&8/64&72/64&648/64\\ 10&7/64&70/64&700/64\\ \vdots\\ 16&1/64&16/64&256/64\\ \hline \text{total}&&9&91.5\\ \end{array}$$Hence:$$\mu_{\bar{X}}=E(\bar{X})=\sum XP(X)=9;\\ \sigma_{\bar{X}}=\sqrt{E(X^2)-[E(X)]^2}=\sqrt{91.5-9^2}=\sqrt{10.5}\approx 3.24.$$It confirms the previous answer.
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