excans:647f0de55f: Difference between revisions
From Stochiki
(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</...") |
mNo edit summary |
||
| Line 1: | Line 1: | ||
'''Solution: E''' | '''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. | 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>0.2n</math> and variance <math>n\sigma^2</math>. Hence the probability of a net loss equals <math display = "block">P(Z \geq \frac{-0.2n}{\sigma \sqrt{n}}) = P(Z \geq \frac{-\sqrt{n}}{3})</math> where <math>Z</math> is a standard normal variable. The 1<sup>th</sup> percentile for a standard normal equals -2.326, hence we need | ||
<math display = "block"> | <math display = "block"> | ||
\frac{\sqrt{n}}{ | \frac{\sqrt{n}}{3} \geq 2.326 \implies n \geq 48.69 | ||
</math> | </math> | ||
Latest revision as of 13:30, 28 November 2024
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]0.2n[/math] and variance [math]n\sigma^2[/math]. Hence the probability of a net loss equals
[[math]]P(Z \geq \frac{-0.2n}{\sigma \sqrt{n}}) = P(Z \geq \frac{-\sqrt{n}}{3})[[/math]]
where [math]Z[/math] is a standard normal variable. The 1th percentile for a standard normal equals -2.326, hence we need
[[math]]
\frac{\sqrt{n}}{3} \geq 2.326 \implies n \geq 48.69
[[/math]]