# Math Help - Central Limit Theorem with exp. distribution

1. ## Central Limit Theorem with exp. distribution

A random sample of size 16 is taken from an exponential distribution, with mean 2.

a) Even though the sample size is not large enough, use the CLT to estimate the
probability that the sample mean is greater than 2.

b) Compute the exact probability that the sample mean is greater than 2.

c) Repeat (a) and (b), for a sample of size 64. Is your approximation closer than it
was for a sample of size 16?

How do you find sample mean and sample variance to plug into the CLT?

2. Originally Posted by wolverine21
A random sample of size 16 is taken from an exponential distribution, with mean 2.

a) Even though the sample size is not large enough, use the CLT to estimate the
probability that the sample mean is greater than 2.

b) Compute the exact probability that the sample mean is greater than 2.

c) Repeat (a) and (b), for a sample of size 64. Is your approximation closer than it
was for a sample of size 16?

How do you find sample mean and sample variance to plug into the CLT?
In each case you're interested in the distribution of the sample mean, that is, the distribution of $\bar{X} = \frac{\sum_{i=1}^{16} X_i}{n}$ where each $X_i$ follows an exponential distribution with mean 2.

In (a) you want an approximation to the distribution of $\bar{X}$. In (b) you need the exact distribution.

a) The Central Limit Theorem says that the limiting distribution of $Z_n = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$ is Normal(0, 1).

In your case $\mu = 2$, $\sigma = 2$ and n = 16.

Use the above approximation to calculate $\Pr(\bar{X} > 2) = \Pr(Z > 0)$. (You don't really need to do any calculation here because the answer is obviously 0.5).

b) Calculate the exact distribution of $\bar{X}$ by following the outline given in this thread: http://www.mathhelpforum.com/math-he...a-samples.html NB: $X_i$ follows a different distribution in that thread but the idea is the same.

Use the pdf of $\bar{X}$ to calculate the exact value of $\Pr(\bar{X} > 2)$.