Central Limit Theorem & Gamma Distribution

[htdc]

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]

Hi there,

Welcome to MHB :)

The central limit theorem states that for a sufficiently large $n$ the value of [math]\frac{S_n-n \mu}{\sigma \sqrt{n}}[/math] 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

 

[htdc]

Hi there,

Welcome to MHB :)

The central limit theorem states that for a sufficiently large $n$ the value of [math]\frac{S_n-n \mu}{\delta \sqrt{n}}[/math] 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]

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]

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

$\chi$ $\sigma$ View attachment 448