If a random sample of size 16 is taken with replacement from the population 1, 1, 1, 2, 4, 5, 5 and 6, what is the mean, variance and standard deviation of the sampling distribution?

Question

Find the following:

* MEAN OF SAMPLING DISTRIBUTION

* STANDARD DEVIATION OF THE SAMPLING DISTRIBUTION

* VARIANCE OF THE SAMPLING DISTRIBUTION

Answer 1

A sampling distribution is defined over ALL possible samples of size 16. If the sample size were more manageable, we would list ALL possible samples of size 16, but there are 8 elements to choose from, making 8¹⁶ possible samples (over 281 trillion!) so that is not a method we want to use. Since we are sampling with replacement, we can generate a discrete probability distribution for the distinct values of X: {1,2,4,5,6}. The P(X=1) = 3/8, P(X=2)= 1/8, P(X=4) = 1/8, P(X=5) = 2/8, P(X=6) = 1/8, or in tabular form,

X P(X) X•P(X) X² X²•P(X)

1 0.375 0.375 1 0.375

2 0.125 0.250 4 0.500

4 0.125 0.500 16 2.000

5 0.250 1.250 25 6.250

6 0.125 0.625 36 4.500

total 1.000 3.125 total "sum"

The mean of a discrete random variable is µ = ∑ (x • p(x)) = 3.125

The standard deviation is σ = Σ (x² • p(x)) - µ² = [you can fill this in now!] = "sum" - 3.125²

The variance is standard deviation squared.