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Let $X$ have gamma distribution with parameters $\alpha=7$ and $\lambda=1$. Investigate the value of $F_X(10)$ using these methods:

  • Find a lower bound using Chebychev's inequality.
  • Approximate the value using the central limit theorem.
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    Well, if W_i is the waiting time for the ith arrival, and X is the gamma random variable, X is the sum of w_i with i=1,2,...,7 but I'm still not sure how to use this2012-11-30

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The central limit theorem says that the average of iid random variables $y_i$ having mean $\mu$ and variance $\sigma^2$ converges to a normal distribution with mean $\mu$ and variance $\sigma^2/n$.

You know that we can express the gamma-distribution random variable $X$ as $X = \sum_{i=1}^7 W_i $ where $W_i \sim \operatorname{Exp}(1)$ has mean 1 and variance 1. That's a sum, not an average, but we can divide both sides by $7$ and then apply the CLT: $X/7 = \frac{1}{7}\sum_{i=1}^7 W_i \; \dot\sim \; \operatorname{Normal}(1,1/7).$ So, multiplying by 7 again, $X \; \dot\sim \; \operatorname{Normal}(7,7).$ Now you can use the normal CDF to approximate $F_X(10)$.

Note that we could have decomposed the Gamma into a sum of $k$ iid $\operatorname{Gamma}(7/k,1)$ random variables for any $k \in \mathbb{N}$ (this property of a distribution is called "infinite divisibility"). We could have ended up with a sum of 16 billion $W_i$ instead of just 7, but our end result would have been exactly the same. This illustrates the point that there isn't a magic number (30, often) after which the CLT "applies."

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    Thanks for catching that the mean should be $7$, not $1$, but it should be a $\operatorname{Normal}(7,7)$, not $(7,1)$. That there's no guarantee the approximation is good is exactly the point I was making in my last paragraph, but this is the approximation the problem asked for.2012-12-03