this in r language The following code is a simulation of the central limit theorem. In the code we generate 36 observations from a random exponential distribution with mean 1 and variance 1. We then use a for loop to generate averages of these 36 observations to display in a histogram. Run the following code in R and select all answers that apply based on the code. I would also suggest running the code more than one as the numbers generated are random and results will vary slightly. CODE: #Generating data data.exp = rexp(36) par(mfrow=c(2,1)) hist(data.exp, main = 'Histogram of exponential with mean 1')
this in r language
The following code is a simulation of the central limit theorem. In the code we generate 36 observations from a random exponential distribution with mean 1 and variance 1. We then use a for loop to generate averages of these 36 observations to display in a histogram.
Run the following code in R and select all answers that apply based on the code. I would also suggest running the code more than one as the numbers generated are random and results will vary slightly.
CODE:
#Generating data
data.exp = rexp(36)
par(mfrow=c(2,1))
hist(data.exp, main = 'Histogram of exponential with mean 1')
#Gererate averages for the central limit theorem
#I.e. data set of averages
bar = c(1:500)#store the averages
for (i in c(1:500)) {
data.exp = rexp(36)
bar[i] = mean(data.exp)
}
hist(bar, main = 'Histogram of data set of averages for CLT')
sprintf("Theoretical mean of data.exp: %s", 1)
sprintf("mean of the average: %s", mean(bar))
sprintf("Theoretical variance of data.exp: %s", 1)
sprintf("variance of the average: %s", var(bar))
![The original data set is normally distributed.
The data set of averages is NOT approximately normally distributed.
The variance of the exponential is approximately equal to the variance of the averages.
The mean of the exponential is NOT approximately equal to the mean of averages the
The mean of the exponential is approximately equal to the mean of averages the
The variance of the averages is approximately equal to the variance of the exponential divided by N.
The original data set is NOT normally distributed
The data set of averages is approximately normally distributed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09fcf446-1afb-4b15-acb5-39347ac88924%2Fa103db9d-8b71-4f9d-adfd-95ad1ea60bf9%2Fbuvi16p_processed.png&w=3840&q=75)
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