# Thread: help with exponential distribution

1. ## help with exponential distribution

X1,X2,...,XN are independently identically exponentially distributed with expected value of 5. How can I compute X[bar]n when n=20 and N=1000? Then compute the proportion of values of X[bar]n that lie between 6.99 and 7.01.

repeat the above question with n=100

My thoughts
so basically i am using code in R software to do this
and basically this question means that suppose there are 1000 iid exp. dist. with Expected value of each X is 5. Then get the mean of each X where the number of observations is 20 then 100.
i used
a=(1:1000) <--makes a vector [1,2,3,...,1000]
for(i in a){a[i]=mean(rexp(20,5))} <---so 20 observations and expected value of 5, each slot in the vector gets replaced by a mean
plot(a)
and
a=(1:1000)
for(i in a){a[i]=mean(rexp(100,5))}
plot(a)
i get the y-axis to be 0.1 to 0.35 for n=20 and 0.14 to 0.26 for n=100, what am I doing wrong? I dont get how to get a proportion of values of X[bar]n that lie between 6.99 and 7.01.

2. ## Re: help with exponential distribution

Originally Posted by omega
X1,X2,...,XN are independently identically exponentially distributed with expected value of 5. How can I compute X[bar]n when n=20 and N=1000? Then compute the proportion of values of X[bar]n that lie between 6.99 and 7.01.
..... I dont get how to get a proportion of values of X[bar]n that lie between 6.99 and 7.01.
Obviously, one way of doing this (obtain the proportion) would be to use a normal approximation...Is this the intent of the assignment you're working on?...Or is it something else?

3. ## Re: help with exponential distribution

It didn't say anything about using normal approximation but how would you do it if I had to?

4. ## Re: help with exponential distribution

Originally Posted by omega
It didn't say anything about using normal approximation but how would you do it if I had to?
It's easy because we know that the mean is 5 and therefore the standard deviation will also be 5.

Thus, we have

6.99 = z*(5/Sqrt[n]) + 5

7.01 = z*(5/Sqrt[n]) + 5.

Just solve for z in both cases and use the normal curve to determine the proportions between the two points.

5. ## Re: help with exponential distribution

so the formula is
x = (mu*z) / sqrt(n) + mu

so then i get

(x-mu)sqrt(n) / mu = z

so

(6.99-5)sqrt(20) / 5 = z
z =~1.77991011
(7.01-5)sqrt(20) / 5 = z
z =~1.797795654

(6.99-5)sqrt(100) / 5 = z
z =3.98
(7.01-5)sqrt(100) / 5 = z
z =4.02

So here i just get the area under the curve between the two points for each one, from the *standard normal distribution* ?

6. ## Re: help with exponential distribution

Originally Posted by omega

So here i just get the area under the curve between the two points for each one, from the *standard normal distribution* ?
Yes, the standard normal distribution.

7. ## Re: help with exponential distribution

so for n=20 i get 0.0014406382
and for n=100 i get 0.0000053585

Does this seem right?

8. ## Re: help with exponential distribution

Originally Posted by omega
so for n=20 i get 0.0014406382
and for n=100 i get 0.0000053585

Does this seem right?
Yes, I checked your calculation for n=100, and it is correct.

However, note that this is a normal approximation to the distribution of the sample means.

I also checked empirically (based on n=100 columns with 1000000 rows) for n=100 and I obtained a proportion of 0.00065. So the normal approximation is going to be a bit off.

The reason is because the sampling distribution of the mean will still be somewhat positively skewed and have some kurtosis. More precisely, the skew of the sampling distribution will be 2/Sqrt[n] and the kurtosis will be 6/n.

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