OmCheeto said:
[...]
View attachment 318330
spreadsheet generated (Mac 'Numbers')
Btw, can someone check my sigma and mu numbers. [...]
You can calculate the \mu and \sigma discretely using the following formulas:
\mu = \frac{\sum x_n}{N}
\sigma = \sqrt{\frac{\sum \left( x_n - \mu \right)^2}{N-1}}
So let's do each individually.
Wordlebot:
\mu = \frac{9(3)+ 9(4)+ 3(5)}{21} = \frac{78}{21} \approx 3.714
\sigma = \sqrt{ \frac{9 \left( 3 - 78/21 \right)^2 + 9 \left(4 - 78/21 \right)^2 + 3 \left( 5 - 78/21 \right)^2 }{20}} \approx 0.7171
OmCheeto:
\mu = \frac{3(2) + 10(3) + 6(4) + 2(5)}{21} = 70/21 \approx 3.333
\sigma = \sqrt{\frac{ 3 \left(2 - 70/21 \right)^2 + 10 \left( 3 - 70/21 \right)^2 + 6 \left(4 - 70/21 \right)^2 + 2 \left( 5 - 70/21 \right)^2}{20}} \approx 0.8563
(All this assumes that I didn't mess up the arithmetic myself.)
The bell curve might not be the best probability density function for your guesses.
What you can say though, according to the Central Limit Theorem, if you average your guesses -- let's say you average them over k trials, the probability density of your
averaged guesses will approach the shape of the bell curve, AND, \mu_k \approx \mu, and \sigma_k \approx \frac{\sigma}{\sqrt{k}}, and these approximations will become more precise for increasing k.
[Edit: Oops. Forgot to take a square-root calculation on OmCheeto's results. Correction made.]
