R codes
> Step 1
>
> random_sample = sample(1:5,1)
> sample = rep(random_sample,147)
> length(sample)
[1] 147
>
>
> Step 2
>
> mean = mean(sample)
> variance = var(sample)
>
> Step3
>
> mean1 = variance1 = c()
> for (i in 1: 9999)
+ {
+ random_sample = sample(1:5,1)
+ sample = rep(random_sample,147)
+ mean1[i] = mean(sample)
+ variance1[i] = var(sample)
+ }
>
a)
From this we can say that all numbers are drawn in equal proportion.
b)
Expected value of mean of all 10000 values of mean is (1+2+3+4+5)/5 = 3
From sample Value of mean for 10000 values of mean is 2.9999
Expected value of variance of all 10000 value of mean is var of 1,2,3,4,5 which is 2.5
From sample Value of variance for 10000 values of mean is 2.013
c)
Shape of histogram has uniform distribution. All points between 1 and 5 are uniformly distributed.
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