The random variables X; are i.i.d. with mean 2 and standard deviation equal to 3. Assume that the Xi are nonnegative. Let Use the CLT to find good approximations to the following quantities. Please find a) P(S100<= 245) b) We let N (a random variable) be the first value of n for which Sn exceeds 119. P(N > 49) ~ c) What is the largest possible value of n for which we have P(Sn <= 128) ~ 0.5?
a) First, we calculate the mean and variance of the sum S100:
mean of S100 = n * mean of X = 100*2 = 200
variance of S100 = n * variance of X = 100*3^2 = 900
standard deviation of S100 = sqrt(variance of S100) = sqrt(900) = 30
Now we standardize the sum S100 to use the CLT:
Z = (S100 - mean of S100) / std deviation of S100
Z = (245 - 200) / 30
Z = 45 / 30
Z = 1.5
Using a standard normal table, we find P(Z <= 1.5) = 0.9332
Therefore, P(S100 <= 245) ≈ P(Z <= 1.5) = 0.9332
b) We are looking for the probability that the sum S_n exceeds 119 for the first time at n > 49. This is equivalent to finding the complement of the probability that the sum does not exceed 119 for the first 49 values:
P(N > 49) = P(S1 <= 119, S2 <= 119, ..., S49 <= 119)
Since the Xi's are nonnegative, P(Sn <= 119) = 0.5
Therefore, P(N > 49) ≈ 0.5^49 = 7.88 x 10^-16
c) We want to find the largest value of n such that P(S_n <= 128) ≈ 0.5
Similar to part a), we standardize the sum Sn:
mean of Sn = n * mean of X = 2n
standard deviation of Sn = sqrt(n) * std deviation of X = sqrt(n)*3
Z = (128 - 2n) / (3*sqrt(n))
Setting Z = 0, we get:
0 = (128 - 2n) / (3*sqrt(n))
128 = 2n
n = 64
Therefore, the largest possible value of n for which P(Sn <= 128) ≈ 0.5 is n = 64.