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I'm trying to solve a pretty straight forward problem but i can't find good info on the subjects necessary to solve it so i'm terribly stuck. I'll present it as follows and later try to explain my specific doubts about it. Here it goes:

Let $X_{1}, \ldots , X_{15}$ be a simple sample from a exponential population $X$ with mean $= 3$ and let $S = X_{1} + \cdots + X_{15}$. Find $P(S>38)$ using:

  • The exact distribution of $S$

  • The normal approximation. In this case, explain why the of the central limit theorem is valid

First:I can't find good information about the exact distribution, so i don't know where should I put the values from the problem, because I don't even know the pdf of the exact distribution

Second I understand the intuition of the CLT in which given a sufficiently large sample size from a population (with a finite level of variance), the mean of all samples from the same population will be approximately equal to the mean. BUT i can't relate that to the problem, it simply doesn't make sense to me how one connects to another. I know about the normal distribution, but what's the meaning of a 'normal approximation' ?

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Your paragraph labeled "second" shows you don't know what the CLT says. The CLT says there's a certain "normal approximation". The normal approximation to the distribution of the sum $S$ is the normal distribution whose expected value and variance are the same as those of $S$. What the CLT says is that the sum $S$ will be approximately normally distributed because $15$ is a large number and $X_1,\ldots,X_{15}$ are independent and all have the same distribution.

You have $\mathbb E(X_1)=3$ and $\operatorname{var}(X_1)=3^2=9$.

Therefore $\mathbb E(S)=15\cdot 3 = 45$ and $\operatorname{var}(S)=15\cdot9=135$.

So $\mathbb E\left(\dfrac{S-45}{\sqrt{135}}\right)=0$ and $\operatorname{var}\left(\dfrac{S-45}{\sqrt{135}}\right)=1$, and $\dfrac{S-45}{\sqrt{135}}$ is approximately normally distributed.

That is what the CLT says about $S$.

Consequently $$ \Pr(S>38)=\left(\frac{S-45}{\sqrt{135}} >\frac{38-45}{\sqrt{135}}\right) = \Pr(Z> -0.602464\ldots), $$ and you get that number from your software or from a table in the back of the book.

The distribution of $X_1$ is $e^{-x/3}\left(\dfrac{dx}{3}\right)$ for $x>0$.

(The density is $e^{-x/3}\cdot\frac 1 3$ for $x>0$; the cumulative distribution function you get by integrating the thing above, and is one of the simplest expressions involved in this cases: $1-e^{-x/3}$ for $x>0$ (and $0$ for $x\le 0$).)

The distribution of $S$ is a Gamma distribution: $$ \text{constant}\cdot x^{15-1} e^{-x/3}\left(\frac{dx}{3}\right). $$

To find the constant, you can write $$ \begin{align} \int_0^\infty x^{15-1} e^{-x/3}\left(\frac{dx}{3}\right) = & 3^{15-1} \int_0^\infty \left(\frac x 3\right)^{15-1} e^{-x/3}\left(\frac{dx}{3}\right) \\[10pt] = & 3^{15-1} \underbrace{\int_0^\infty u^{15-1} e^{-u}\, du}_{\text{a familiar integral}} = 3^{15-1}\cdot(15-1)!\,. \end{align} $$

PS: You misused the word "sample" in the usual way done by people whose background is not in statistics. $X_1,\ldots,X_{15}$ are not $15$ samples; they are $15$ observations in one sample.

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  • $\begingroup$ Here i am in a one man standing ovation for such a beautiful, clear, concise answer. Thanks Michael, that was perfect. $\endgroup$
    – matt_zarro
    Mar 26 '14 at 2:39

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