According to the central limit theorem, as n approaches infinity, the distribution of the random variable S_n approaches a normal distribution with mean zero and variance n*sigma^2.
Therefore, we can standardize the random variable S_n as follows:
Z_n = (S_n - 0) / sqrt(n*sigma^2) = S_n / sqrt(n*sigma^2).
We want to find the probability that S_n is less than or equal to 2*sigma*sqrt(n), which can be written as:
P(S_n <= 2*sigma*sqrt(n)) = P(S_n / sqrt(n*sigma^2) <= 2*sigma*sqrt(n) / sqrt(n*sigma^2))
= P(Z_n <= 2*sigma / sqrt(n*sigma^2))
= P(Z_n <= 2 / sqrt(n)).
As n approaches infinity, 2 / sqrt(n) approaches 0. Therefore, as n approaches infinity, P(Z_n <= 2 / sqrt(n)) approaches Phi(0), which is equal to 0.5.
So, as n approaches infinity, P(S_n <= 2*sigma*sqrt(n)) converges to Phi(0), where a = 0.
As for the probability P(S_n <= 0), since the random variables X_n have mean zero, the sum S_n will also have mean zero. Therefore, as n approaches infinity, P(S_n <= 0) approaches Phi(0), which is equal to 0.5.
So, as n approaches infinity, P(S_n <= 0) converges to 0.5.
Let X_ n be i.i.d. random variables with mean zero and variance \sigma ^2. Let S_ n=X_1+\cdots +X_ n. Let \Phi stand for the standard normal CDF. According to the central limit theorem, and as n\to \infty, \mathbf{P}(S_ n\leq 2\sigma \sqrt{n}) converges to \Phi (a), where:
a=\,
Furthermore,
\mathbf{P}(S_ n\leq 0) converges to:
(Here, enter the numerical value of the probability.)
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