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| rev | Admin | (Created page with "'''Solution: E''' The expected number of pennies in a single roll equals <math>\mu = 49.8 </math> and the variance equals <math>\sigma^2 = 0.36 </math>. In particular, the net loss for <math>n</math> rolls is approximately normally distributed with mean <math>0.02n</math> and variance <math>n\sigma^2</math>. Hence the probability of a net loss equals <math display = "block">P(Z \geq \frac{-0.02n}{\sigma \sqrt{n}}) = P(Z \geq \frac{-\sqrt{n}}{30})</math> where <math>Z</...") | Jun 28'24 at 1:26 | +683 |