1. The problem statement, all variables and given/known data How can I derive the probability density function by using the Central Limit theorem? For an example, let's say that we have a random variable Xi corresponding to the base at the ith position; to make even simpler, let's say all probabilities are equal. If we have four variable, P(Xi = A) =P (Xi = B) = P(Xi = C) = P(Xi = D) the probability density function would be fX(x) =(1/√2πσ)e^[-(x−µ)^2/2σ^2]= (1/√2π)e^(-x^2/2) since µ = 0 and σ = 1. Plus we would have a probability of 1/4 for all i. Now consider random strands of length l, where l is very large. How can I use Use Central Limit Theorem to find the probability density function corresponding to finding the length N in A(P(Xi = A)) occurrence times? 2. Relevant equations fX(x) =(1/√2πσ)e^[-(x−µ)^2/2σ^2]= ∅(x)=(1/√2π)e^(-x^2/2) when µ = 0 and σ = 1 3. The attempt at a solution The theorem says : "Let X1, X2, . . . , Xn , . . . be a sequence of independent discrete random variables, and let Sn = X1 + X2 + · · · + Xn. For each n, denote the mean and variance of Xn by µn and σ(^2)n, respectively. Define the mean and variance of Sn to be mnand s^(2)n, respectively, and assume that sn → ∞. If there exists a constant A, such that |Xn| ≤ A for all n, then for a < b, limn→∞P (a < Sn − mn/sn< b)=1/√2π∫(from a to b)e^−x(^2)/2dx" Well lets say XA be the random variable corresponding to the number of oc- curences of the base A in the strand. My guessed would be that the mean length for every occurrence of XA, results in a mean length of 4 bases. Because bases are equally likely, then Xa occurs in roughly 1/4 of the positions. Therefore variance would be 16 and all I would do is plug it in fX(x) =(1/√2πσ)e^[-(x−µ)^2/2σ^2] However how do I use the Central Limit theorem to find p(x)?