- #1
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.
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.