00:01
Here we are going to find the euclidean distance between the observations 10 and 4, that is d10 comma 4, which is equal to square root of 0 .032 minus 0 .510 the whole square plus 0 .741 minus 0 .207 the whole square plus 10 .7 minus 0 .207 the whole square plus 10 .7, plus 0 .004 the whole square.
00:34
Square here it is plus plus minus 0 .892 plus 0 .2 .0 .219 the whole square plus minus 0 .17340 plus 0 .943 the whole square plus 0 .693 plus 0 .40 plus 0 .602 the whole square plus 1 .62 plus 1 .692 plus 1 .62 28 the whole square plus minus 0 .863 plus 0 .724 the whole square.
01:15
Everything comes under the square root.
01:17
Upon simplification, we will get the value for this as 1 .4916, which is approximately 1 .492.
01:25
Next we have to find eukudian distance between the observations 10 and 20.
01:30
It is d of 10 .20, which is equal to square root of 0 .032 minus 0 .4 double 6 the whole square plus 0 .741 minus 0 .474 the whole square plus 0 .7 plus 0 .490 the whole square plus minus 0 .892 minus 0 .892 minus 0 .65 the whole square plus minus 0 .173 minus 0 .083 the whole square plus minus 0 .693 plus 0 .458 the whole square plus 1 .62 minus 1 .62 minus 1 .7 double 3 the whole square plus minus 0 .863 plus 0 .721 the whole square.
02:30
So which is equal to upon simplification we will get the 3 .63.
02:34
Value for this as 2 .055.
02:36
Next, the euclidean distance between the observations, 13 and 20, we have to find it out.
02:43
So d -13 comma 20, which is equal to square root of 0 .195 minus 0 .4 -6 the whole square plus 0 .875 minus 0 .474 the whole square plus 0 .748 plus 0 .748 plus 0 .4 .5 plus 0 .748 plus 0 .4...