00:02
Hello students, let's do this question.
00:03
First of all, let's do the a part here.
00:06
So here, the rth moment of a variable x about the mean x bar is denoted by mu, r.
00:39
And the formula for mu of r is equal to 1 upon an summation i.
00:51
X i minus x bar multiply by r so for the first part here n is equal to number of observations that is 10 here so mean x bar is 73 .8 now the first moment therefore r is equal to 1 so here mu 1 will be equal to to 1 upon 10 95 minus 73 .8 plus and similarly up to 58 minus 73 .8.
01:50
So by solving we will get our mu 1 as 0.
01:56
Moving forward here second moment therefore r is equal to 2.
02:10
So mu 2 is equal to 1 upon n summation i x i x i minus x bar so here our mu two will be equal to 1 upon 10 here 95 minus 73 .8 square plus 67 minus 73 .8 square and up to 58 minus 73 .8 square and up to 58 minus 73 .8 square.
03:02
So by solving this we will get mu 3 as 84 .0 to 4.
03:13
Now moving further here we have to calculate 4th moment.
03:28
So 4th moment mu 4 by using the same formula will be equal to 54.
03:44
028 .931 here.
03:47
Now let's move to the b part of the question here.
03:53
We have to calculate beta 1 and beta 2.
04:00
Beta 1 is equal to mu 3 square upon mu 2 square...