Comparing Independent Random Flips: Normal vs. Binomial Distribution

candyduz
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Which of the following is more likely to be generated using independent random flips? Explain.


table.jpg


Pls help! Thanks!
 
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What have you tried?

Where are you stuck?
 
I was wondering if it's something to do with normal distribution or central limit theorem but honestly, I have no idea at all on how to solve this question
 
I tried to approximate normal to binomial and out of groups of 20 flips, for the first set, I got 10, 6, 9, 8, 8, 11, 11, 10, 9, 11 "0"s and for the second set, I got 12, 9, 9, 9, 10, 12, 10, 12, 11, 11 "0"s.

Am I on the right track?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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