Help with exponential distribution

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SUMMARY

The discussion revolves around calculating the sample mean \( \bar{X}_n \) from 1000 independent and identically distributed (iid) exponential random variables with an expected value of 5. The user employs R software to compute the means for sample sizes of 20 and 100, using the function `rexp()`. The user seeks clarification on how to determine the proportion of sample means that fall within the range of 6.99 to 7.01. Key insights include the relationship between the mean and variance of the exponential distribution, which is essential for understanding the distribution of sums of iid variables.

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sneaky666
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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=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=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 don't get how to get a proportion of values of X[bar]n that lie between 6.99 and 7.01.
 
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sneaky666 said:
and basically this question means that suppose there are 1000 iid exp. dist. with Expected value of each X is 5.

Correct, and before you start writing any code, think about what that statement implies.

For an exponential distribution, if you are told the mean you also know the variance.

What general results do you know about the distribution of the sum of independent identically distributed variables?
 

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