Answer to Question #239085 in Statistics and Probability for Adriano

Question

a large population consist of equal numbers of the digits 1 and 3. (a) Find the mean and variance of this population (b) Find the probability distribution of the mean of samples size three taken from this population and verify that this mean mean is equal to the population mean and its variance is equal to one-third of the  population variance

Answer 1

Solution:

Let there be 'n' numbers of digit 1 and so 'n' numbers of digit 3.

Total number of numbers in population = n+n = 2n.

(a) Mean$=\bar x=\dfrac{Sum\ of \ observations}{Total\ no. \ of \ observations}$

$=\dfrac{1n+3n}{2n}=2$

Variance$=\sigma^2=\dfrac{\Sigma fx^2}{\Sigma f}-(\bar x)^2=\dfrac{n(1^2)+n(3^2)}{2n}-2^2=1$

(b) Samples of size 3 are (1,1,1),(3,3,3),(1,1,3),(1,3,3)

Sample Mean$=\dfrac{\dfrac{1+1+1}{3}+\dfrac{3+3+3}{3}+\dfrac{1+1+3}{3}+\dfrac{1+3+3}{3}}4=2$, which is same as population mean.

Sample variance$=\dfrac{N-n'}{N-1}\times \dfrac{\sigma^2}{n'}$

Here, N=2n, and n'=3

So, Sample variance$=\dfrac{2n-3}{2n-1}\times \dfrac{1}{3}=\dfrac{1}{3}$ [when n is large]

which is 1/3 of population variance.

Hence, verified.