Question

17. A variable takes only two distinct values a and b, with same frequency. Find the
second, third and fourth central moments.​

Answer 1

Answer:

sorry I don't know the answer

Answer 2

Given two distinct values of a variable with same frequency each, find the second , third and fourth central moments

Explanation:

1. let the variable be 'x' and the frequency of both values of x- 'a' and 'b' be 'n' the, $\mu_1=\sum xP(x)=\frac{na+nb}{2n}\ \ \ \ \ =>\mu_1=\frac{a+b}{2}$          
2. now second , third , fourth moments about origin are given by,                          $\mu_2=\sum x^2P(x)=\frac{na^2+nb^2}{2n} \ \ \ \ \ => \mu_2=\frac{a^2+b^2}{2}$      
3. similarly, $\mu_3=\frac{a^3+b^3}{2} \ \ \ \ \ \mu_4=\frac{a^4+b^4}{2}$  
4. now central moments using moments about origin are given by,                            $\mu_2'=\mu_2-\mu_1^2 \\\mu_3'=\mu_3-3\mu_2\mu_1+2\mu_1^3\\\mu_4'=\mu_4-4\mu_3\mu_1+6\mu_2\mu_1^2-3\mu_1^4$      ---(a)
5. hence using above values in (a) we get,                                                                              $\mu_2'=\frac{a^2+b^2}{2}-\frac{(a+b)^2}{4}= \frac{(a-b)^2}{4}\\\mu_3'= \frac{a^3+b^3}{2}-3(\frac{a^2+b^2}{2} )(\frac{a+b}{2} )+2(\frac{(a+b)^3}{8} ) =0\\\mu_4'= \frac{a^4+b^4}{2}-4(\frac{a^3+b^3}{2} )(\frac{a+b}{2} )+6(\frac{a^2+b^2}{2} )(\frac{(a+b)^2}{4} ) -3(\frac{(a+b)^4}{16}) = \frac{(a-b)^4}{16}$  
6. hence second, third and fourth central moments are $\frac{(a-b)^2}{4}, 0\ and\ \frac{(a-b)^4}{16}$ respectively