For questions 3 to 5 assume the data are SAMPLES taken from

For questions 3 to 5, assume the data are SAMPLES taken from a large population. (MAKE ATABLE for your answers. Standard Mean Median Sum of Squares t Variance Deviation Question Data (a sample) 0.31 0.53 0.64 3.1 0 -2.4 4.1 6. For the data of problem #5 (Question #5) above, calculate the following two items: a.) Coefficient of Variation b.) Coefficient of Dispersion

Solution

5.)

6.) Coefficient of variation = std deviation/ mean

Coefficient of dispersion = variance/mean

0.1372/0.4933 = 0.2781

Sample	mean	median	sum of squares	variance	Standard deviation
0.31,0.53,0.64	(0.53 + 0.31 + 0.64)/3 = 0.4933	0.53	(0.31 - 0.4933)^2 + (0.53 - 0.4933)^2 + (0.64 - 0.4933) ^2 = 0.0565	(0.0565/3) = 0.01882	0.01882 = 0.1372
-2.2,-3.1 ,- 4.8	(-2.2 + -3.1 + -4.8)/3 = -3.3667	-3.1	(-2.2 + 3.3667)^2 + (-3.1 - 3.3667)^2 + (-4.8 - 3.3667)^2 = 3.48667	(3.48667/3) = 1.1622	1.1622 = 1.0781
3.1,0,-2.4,4.1	(3.1 + 0 - 2.4 + 4.1)/4 = 1.2	(0 + 3.1)/2 = 1.55	(3.1 - 1.2)^2 + ( 0 - 1.2)^2 + (-2.4 - 1.2)^2 + ( 4.1 - 1.2)^2 = 26.42	(26.42/4) = 6.605	6.605 = 2.5700