Question -
Answer -
From the given data,
Let us make the tableof the given data and append other columns after calculations.
| X (xi) | Y (yi) | Xi2 | Yi2 |
| 35 | 108 | 1225 | 11664 |
| 54 | 107 | 2916 | 11449 |
| 52 | 105 | 2704 | 11025 |
| 53 | 105 | 2809 | 11025 |
| 56 | 106 | 8136 | 11236 |
| 58 | 107 | 3364 | 11449 |
| 52 | 104 | 2704 | 10816 |
| 50 | 103 | 2500 | 10609 |
| 51 | 104 | 2601 | 10816 |
| 49 | 101 | 2401 | 10201 |
| Total = 510 | 1050 | 26360 | 110290 |
We have to calculateMean for x,
Mean x̅ = ∑xi/n
Where, n = number ofterms
= 510/10
= 51
Then, Variance for x =
= (1/102)[(10× 26360) – 5102]
= (1/100) (263600 –260100)
= 3500/100
= 35
WKT Standard deviation= √variance
= √35
= 5.91
So, co-efficient ofvariation = (σ/ x̅) × 100
= (5.91/51) × 100
= 11.58
Now, we have tocalculate Mean for y,
Mean ȳ = ∑yi/n
Where, n = number ofterms
= 1050/10
= 105
Then, Variance for y =
= (1/102)[(10× 110290) – 10502]
= (1/100) (1102900 –1102500)
= 400/100
= 4
WKT Standard deviation= √variance
= √4
= 2
So, co-efficient ofvariation = (σ/ x̅) × 100
= (2/105) × 100
= 1.904
By comparing C.V. of Xand Y.
C.V of X > C.V. ofY
So, Y is more stablethan X.