Question

the first four Moments a distribution
about x=2, are 1, 2.5, 5.5 and 16 .
calculate the four Moments
(I) about the Mean
(2) about the zero
​

Answer 1

The first four moments about the mean are 0, 1.5, 0, and 6 and the first four moments about zero are 3, 10.5, 5.5, 16

Given:

The first four Moments a distribution about x=2, are 1, 2.5, 5.5, and 16.

To Find:

The first four moments

(I) about the mean

(2) about the zero.

Solution:

Let µ$_{1}$', µ$_{2}$', µ$_{3}$' and µ$_{4}$' be the first four moments about x=2.

So we have

µ$_{1}$' = 1, µ$_{2}$' = 2.5, µ$_{3}$' = 5.5 and µ$_{4}$' = 16 and A = 2

1) We need to find the first four moments about mean, i.e. to find out µ$_{1}$, µ$_{2}$, µ$_{3}$ and µ$_{4}$.

By definition,  µ$_{1}$ = 0

µ$_{2}$ = µ$_{2}$'- µ$_{1}$'^$2$ = 2.5- 1 = 1.5

µ$_{3}$ = µ$_{3}$'-3µ$_{2}$'µ$_{1}$'+ 2µ$_{1}$'^$3$ = 5.5 -3(2.5) +2(1)= 5.5-7.5-2 = 0

µ$_{4}$ = µ$_{4}$'-4µ$_{3}$'µ$_{1}$'+ 6µ$_{2}$'µ$_{1}$'-3 µ$_{1}$'^$4$ = 16-4(5.5 x 1)+ 6(2.5x1)-3 = 16-22 +15-3 = 6

Hence first four moments about mean are 0, 1.5, 0 and 6.

2) Let us find the four moments say v$_{1}$, v$_{2}$, v$_{3}$, and v$_{4}$ about zero.

v$_{1}$ = µ$_{1}$'+A = 1+2 = 3

v$_{2}$ = µ$_{2}$ + v$_{1}$^$2$ = 1.5+9 = 10.5

v$_{3}$ = µ$_{3}$ - 3v$_{2}$v$_{1}$+ 2v$_{1}$^$3$ = 0+3(10.5x3)+2($3^{3}$) = 40.5

v$_{4}$ = µ$_{4}$ - 4v$_{3}$v$_{1}$+ 6v$_{2}$v$_{1}$- 3 v$_{1}$^$4$ = 6 -4( 40.5x3)+6( 10.5x3) = 168

Hence first four moments about zero are 3, 10.5, 5.5, 16.

#SPJ3

Answer 2

Answer:

Step-by-step explanation: