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| revprev | Admin | Jun 28'24 at 18:41 | +9 | m | |
| revcur | Admin | (Created page with "'''Solution: D''' We let <math>X_i = \pm 1 </math> with probability that <math>X_i = 1 </math> equals 1/2. Then we want to approximate the probability that <math>\sum_{i=1}^{100}X_i \in [-10,10] </math>. The expected value of <math>X_i </math> is 0 and its variance is 1. By the central limit theorem, the sum <math>\sum_{i=1}^{100}X_i </math> is approximately normally distributed with mean 0 and variance 100. Hence we have <math display = "block">\sum_{i=1}^{100}X_i \i...") | Jun 27'24 at 19:30 | +599 |