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Suppose that \(X_{i}, i=1,2, \ldots, 20\) are independent random variables, each having a geometric distribution with parameter \(0.8\). Let \(S=X_{1}+\cdots+X_{20}\). Use the central limit theorem \(P(X \geq 18)\).

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The exact numerical answer depends on a custom lookup in the normal distribution table or computation in software.
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Step 1: Identify the Geometric Distribution

Given that \(X_i, i=1,2, \ldots, 20\) are independent random variables each having a geometric distribution with parameter \(0.8\). In this case, the parameter is the probability of success in each trial, denoted as \(p=0.8\). The geometric distribution models the number of failures before the first success.

Step 2: Calculate Mean and Variance

For a geometrically distributed random variable \(X\) with success probability \(p\), the mean (expected value) \(E[X]=(1-p)/p\) and the variance \(Var[X]=(1-p)/p^2\). Here, mean \(μ=(1-0.8)/0.8=0.25\) and variance \(σ^2=(1-0.8)/(0.8)^2=0.3125\).

Step 3: Apply the Central Limit Theorem

We are asked to find \(P(S \geq 18)\), where \(S=X_{1}+\cdots+X_{20}\) represents the sum of 20 independent geometric random variables. We can regard \(S\) as approximately normally distributed because of its large size (20). By the CLT, \(S\) has mean \(E[S]=n*\mu=20*0.25=5\) and variance \(Var[S]=n*σ^2=20*0.3125=6.25\), thus the standard deviation \(σ_S=\sqrt{Var[S]}=\sqrt{6.25}\). Then we standardize \(S\) and change the inequality direction by subtracting \(-18\), giving \(1-P(\frac{S-5}{\sqrt{6.25}} < \frac{18-5}{\sqrt{6.25}})\)

Step 4: Find the Desired Probability

We lookup the right hand side value in a standard normal distribution table or use computational software to find the associated probability. We then subtract this from one to find the required probability \(P(S \geq 18)\).

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