# Normal approximation question?

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• October 20th 2010, 10:01 AM
sankeel
Normal approximation question?
A number is randomly selected from a box with numbers -2, -2, -2, 0, 2, 2 and the number obtained is the amoun of dollar you win or lose

E(x) = -1/3 and Stdv = 1.795

Then after you play 30 games, what is the chance that you will be in will be losing money?
answer is0.8454
• October 21st 2010, 11:46 PM
Sambit
let, $X_i$ = the amount of money I win (or lose) in the $i$th turn. so, after 30 turns, I lose money only if $X_1 + X_2 + X_3 + ...... + X_{30} < 0$ ie, $\sum_{i=1}^{30} X_i < 0$, the probability of which is given by $P(\sum_{i=1}^{30} X_i < 0) = P(\frac{(\sum_{i=1}^{30} X_i) - (30 * -1/3)}{1.795*\sqrt{30}} < \frac{0 - (30 * -1/3)}{1.795*\sqrt{30}}) = \Phi(\frac{0 - (30 * -1/3)}{1.795*\sqrt{30}}) = \Phi(1.01713) = 0.845454$

hence the answer.

note that, if $X_i \sim N(\mu , \sigma^2)$ , then $\sum_{i=1}^{30} X_i \sim N(30\mu , 30\sigma^2)$, where $X_i$'s are i.i.d. variables.
then, by standardization, we get the result.