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Problem 45

# 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.
See the step by step solution

## 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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