Question

It $X$is a gamma random variable with parameters$(n,1)$, approximately how large must $n$be so that$P\left\{\left|\frac{X}{n}\right|-1>.01\right\}<.01?$

Short answer

$n>{258}^2=66564$.

Step-by-step solution

Given Information.

Given$X$is a gamma random variable with parameters$(n,1)$,

Explanation.

Assume that$X$has a gamma distribution with parameters $\alpha =n$and$\lambda =1$. Therefore, the variable$X$is a sum of $n$independent variables $X_i$:

$X=X_1+X_2+\cdots +X_n$

whereby each random variable has an exponential distribution with parameters$\lambda$. The mean and the variance of variables $X$are

$E\left[X\right]=\frac{\alpha }{\lambda }=\frac{n}{1}=n$

and

$Var\left(X\right)=\frac{\alpha }{{\lambda }^2}=\frac{n}{1^2}=n$.

Explanation.

The central limit theorem says that the average of a set of independent identically distributed random variables is approximately normally distributed

for each$a$,

$P\left\{\frac{X-E\left[X\right]}{\sqrt{Var\left(X\right)}}\le a\right\}\to \Phi \left(a\right)$

Now, using this theorem we get:

$$\begin{gathered} .01>P\left\{\left|\frac{X}{n}-1\right|>.01\right\}=1-P\left\{-.01\le \frac{X}{n}-1\le .01\right\}= \\ 1-P\left\{-.01\sqrt{n}\le \frac{X-n}{\sqrt{n}}\le .01\sqrt{n}\right\}\stackrel{(*)}{\approx }1-\left[\Phi \right(.01\sqrt{n})-\Phi (-.01\sqrt{n}\left)\right] \\ \Phi (-z)=1-\Phi \left(z\right)=1-\left[2\Phi \right(.01\sqrt{n})-1]\Rightarrow \Phi (.01\sqrt{n})>.995 \end{gathered}$$

$\stackrel{\text{Table}5.1\text{(textbook, Chapter 5)}}{\Rightarrow }.01\sqrt{n}>2.58\Rightarrow \sqrt{n}>258\Rightarrow n>{258}^2=66564$