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Central Limit Theorem & Gamma Distribution

dcht

New member
Nov 4, 2012
2
The time it takes to complete a project is a random variable Y with the exponential distribution with parameter β=2 hours.

Apply the central limit theorem to obtain an approximation for the probability that the average project completion time of a sample of n=64 projects undertaken independently over the last year will be within 15 minutes of the true mean completion time.

Any ideas on even how to start this? :\
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
Hi there,

Welcome to MHB :)

The central limit theorem states that for a sufficiently large $n$ the value of \(\displaystyle \frac{S_n-n \mu}{\sigma \sqrt{n}}\) is approximately normal. So if you plug in the given information plus make some inferences from the fact that you have an exponential distribution, you can figure this out.

Have you done anything like this already? If you haven't seen it done it's not a process that I think many would just be able to do through intuition.

Jameson
 
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dcht

New member
Nov 4, 2012
2
Hi there,

Welcome to MHB :)

The central limit theorem states that for a sufficiently large $n$ the value of \(\displaystyle \frac{S_n-n \mu}{\delta \sqrt{n}}\) is approximately normal. So if you plug in the given information plus make some inferences from the fact that you have an exponential distribution, you can figure this out.

Have you done anything like this already? If you haven't seen it done it's not a process that I think many would just be able to do through intuition.

Jameson
Thanks for the quick reply. We've done a few similar examples, but like most of our homework, none of the questions match the practice problems.
 

chisigma

Well-known member
Feb 13, 2012
1,704
The time it takes to complete a project is a random variable Y with the exponential distribution with parameter β=2 hours.

Apply the central limit theorem to obtain an approximation for the probability that the average project completion time of a sample of n=64 projects undertaken independently over the last year will be within 15 minutes of the true mean completion time.

Any ideas on even how to start this? :\
A good starting point may be to read the posts...

http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index9.html#post7118

http://www.mathhelpboards.com/f23/unsolved-statistics-questions-other-sites-932/index9.html#post7147

... where is explained that for n 'large enough' the p.d.f. pf the mean of n r.v. each ot them with mean $\mu$ and variance $\sigma^{2}$ is a normal distribution with mean $\mu$ and variance $\displaystyle \sigma^{2}_{n}=\frac{\sigma^{2}}{n}$. In Your case is $\displaystyle \mu=\sigma=\frac{1}{2}$, so that is $\displaystyle \sigma_{n}= \frac{1}{16}$ and the requested probability is...


$\displaystyle P= \text{erf} (\frac{4}{\sqrt{2}})= .99993666575...$

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
The effective computation of erf(x) for x 'large enough' [say x>2.5...] may be a difficult task and in these cases the identity erf(x)= 1-erfc(x) may be sucessfully used. Several years ago I created the annexed table of the erfc(x) function. In this case is $\displaystyle x=\frac{4}{\sqrt{2}} \sim 2.82$ so that is $\displaystyle \text {erfc}\ (x) \sim 7.5\ 10^{-5} \implies \text{erf}\ (x) \sim .999925$...

Kind regards

$\chi$ $\sigma$


log erfc.JPG
 

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