Exercise


ABy Admin
Jun 27'24

Answer

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 \notin [-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]]\sum_{i=1}^{100}X_i \notin [-10,10] \approx P(Z \notin [-1,1])[[/math]]

where [math]Z[/math] is a standard normal. This is approximately equal to 0.3173.

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