Standard variation is an absolute measure of dispersion. When comparison has to be made between two series then the relative measure of dispersion, known as coeff.of variation is used.
Coefficient of Variation, CV is defined and given by the following function:
CV=σX×100CV=σX×100
Where −
· CVCV = Coefficient of Variation.
· σσ = standard deviation.
· XX = mean.
Problem Statement:
From the following data. Identify the risky project, is more risky:
|
Year |
1 |
2 |
3 |
4 |
5 |
|
Project X (Cash profit in Rs. lakh) |
10 |
15 |
25 |
30 |
55 |
|
Project Y (Cash profit in Rs. lakh) |
5 |
20 |
40 |
40 |
30 |
Solution:
In order to identify the risky project, we have to identify which of these projects is less consistent in yielding profits. Hence we work out the coefficient of variation.
|
Project X |
Project y |
||||
|
XX |
Xi−X¯Xi−X¯ |
x2x2 |
YY |
Yi−Y¯Yi−Y¯ |
y2y2 |
|
10 |
-17 |
289 |
5 |
-22 |
484 |
|
15 |
-12 |
144 |
20 |
-7 |
49 |
|
25 |
-2 |
4 |
40 |
13 |
169 |
|
30 |
3 |
9 |
40 |
13 |
169 |
|
55 |
28 |
784 |
30 |
3 |
9 |
|
∑X=135∑X=135 |
|
∑x2=1230∑x2=1230 |
∑Y=135∑Y=135 |
|
∑y2=880∑y2=880 |
Project X
Here X¯=∑XN=∑1355=27and σx=∑X2N−−−−√⇒σx=12305−−−−√=246−−−√=15.68⇒CVx=σxX×100=15.6827×100=58.07Here X¯=∑XN=∑1355=27and σx=∑X2N⇒σx=12305=246=15.68⇒CVx=σxX×100=15.6827×100=58.07
Project Y
Here Y¯=∑YN=∑1355=27and σy=∑Y2N−−−−√⇒σy=8805−−−√=176−−−√=13.26⇒CVy=σyY×100=13.2527×100=49.11Here Y¯=∑YN=∑1355=27and σy=∑Y2N⇒σy=8805=176=13.26⇒CVy=σyY×100=13.2527×100=49.11
Since coeff.of variation is higher for project X than for project Y, hence despite the average profits being same, project X is more risky.