Solution:-
From the given data,
draw the table of the given data and append other columns after calculations.
Now calculate Mean for x,
Mean 
Where, n = number of terms 
Variance for x= ![\frac{1}{{{n}^{2}}}\left[ N\sum x_{1}^{2}-{{\left( \sum {{x}_{i}} \right)}^{2}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c802603d2dd59d8f961f5b119c552f2b_l3.png "Rendered by QuickLaTeX.com")
![=\left( 1/{{10}^{2}} \right)\left[ \left( 10\times 26360 \right)-{{510}^{2}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2860082ff3c5d0b73d2f287d6cbf5aef_l3.png "Rendered by QuickLaTeX.com")
WKT Standard deviation = 
So, co-efficient of variation 
Calculate Mean for y,
Mean 
Where, n = number of terms
Variance for y
![=\frac{1}{{{n}^{2}}}\left[ N\sum y_{1}^{2}-{{\left( \sum {{y}_{i}} \right)}^{2}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5d7445ea04dcbd5a7babcd5b051d0685_l3.png "Rendered by QuickLaTeX.com")
![\left( 1/{{10}^{2}} \right)\left[ \left( 10\times 110290 \right)-{{1050}^{2}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1e39e8cc733c2f616d29a37edd5f1b3c_l3.png "Rendered by QuickLaTeX.com")
WKT Standard deviation 
So, co-efficient of variation
Compare C.V. of X and Y.
C.V of X > C.V. of Y
So, Y is more stable than X.