The variance of the given 9 samples are 0, 4.5, 40.5, 4.5, 0, 18, 40.5, 18, 0 and the probabilities are 3/9, 2/9, 2/9, and 2/9.
It is given that the number of people in the households are 1, 4, and 10.
sample size n = 2 are randomly selected with replacements from the population.
It is required to find the variance of each of the nine samples.
What is the standard deviation?
It is defined as the measure of data disbursement, It gives an idea about how much is the data spread out.
We know the formula for finding the Variance is given below:
[tex]\rm V= \frac{\sum(x_i-X)}{(n-1)}[/tex]
Where V is the Variance
[tex]\rm x_i[/tex] is the value of one sample
X is the mean sample value
n is the sample size.
And the probability = [tex]\rm \frac{Number \ of \ favourable \ outcome}{total \ outcome}[/tex]
For the (1,1) the mean sample value is 1 ( [tex]\rm \frac{1+1}{2} \Rightarrow 1[/tex])
The variance for the sample = [tex]\rm (1-1)^2+(1-1)^2 \Rightarrow 0[/tex]
Similarly calculating variance for all the samples shown in the picture.
Total number of same variance(V=0) = 3
Totall sample = 9
Probabilty = 3/9.
Similarly calculating all the probability for the remaining samples shown in the picture.
Thus, the variance of the given 9 samples are 0, 4.5, 40.5, 4.5, 0, 18, 40.5, 18, 0 and the probabilities are 3/9, 2/9, 2/9, and 2/9.
Learn more about the standard deviation here:
brainly.com/question/12402189
