From the data given below, the state which group is more variable, A or B?
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |
| Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |
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Solution:-
To compare the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.
Co-efficient of variation (C.V.) = (σ/ x̅) × 100
Where, σ = standard deviation, x̅ = mean
For Group A,
| Marks | Group A
fi |
Mid-point
Xi |
Yi = (xi – A)/h | (Yi)2 | fiyi | fi(yi)2 |
| 10 – 20 | 9 | 15 | ((15 – 45)/10) = -3 | (-3)2
= 9 |
– 27 | 81 |
| 20 – 30 | 17 | 25 | ((25 – 45)/10) = -2 | (-2)2
= 4 |
– 34 | 68 |
| 30 – 40 | 32 | 35 | ((35 – 45)/10) = – 1 | (-1)2
= 1 |
– 32 | 32 |
| 40 – 50 | 33 | 45 | ((45 – 45)/10) = 0 | 02 | 0 | 0 |
| 50 – 60 | 40 | 55 | ((55 – 45)/10) = 1 | 12
= 1 |
40 | 40 |
| 60 – 70 | 10 | 65 | ((65 – 45)/10) = 2 | 22
= 4 |
20 | 40 |
| 70 – 80 | 9 | 75 | ((75 – 45)/10) = 3 | 32
= 9 |
27 | 81 |
| Total | 150 | -6 | 342 |

Where A = 45,
and yi = (xi – A)/h
Here, h = class size = 20 – 10
h = 10
So, x̅ = 45 + ((-6/150) × 10)
= 45 – 0.4
= 44.6

σ2 = (102/1502) [150(342) – (-6)2]
= (100/22500) [51,300 – 36]
= (100/22500) × 51264
= 227.84
Hence, standard deviation = σ = √227.84
= 15.09
∴ C.V for group A = (σ/ x̅) × 100
= (15.09/44.6) × 100
= 33.83
Now, for group B.
| Marks | Group B
fi |
Mid-point
Xi |
Yi = (xi – A)/h | (Yi)2 | fiyi | fi(yi)2 |
| 10 – 20 | 10 | 15 | ((15 – 45)/10) = -3 | (-3)2
= 9 |
– 30 | 90 |
| 20 – 30 | 20 | 25 | ((25 – 45)/10) = -2 | (-2)2
= 4 |
– 40 | 80 |
| 30 – 40 | 30 | 35 | ((35 – 45)/10) = – 1 | (-1)2
= 1 |
– 30 | 30 |
| 40 – 50 | 25 | 45 | ((45 – 45)/10) = 0 | 02 | 0 | 0 |
| 50 – 60 | 43 | 55 | ((55 – 45)/10) = 1 | 12
= 1 |
43 | 43 |
| 60 – 70 | 15 | 65 | ((65 – 45)/10) = 2 | 22
= 4 |
30 | 60 |
| 70 – 80 | 7 | 75 | ((75 – 45)/10) = 3 | 32
= 9 |
21 | 63 |
| Total | 150 | -6 | 366 |

Where A = 45
h = 10
So, x̅ = 45 + ((-6/150) × 10)
= 45 – 0.4
= 44.6

σ2 = (102/1502) [150(366) – (-6)2]
= (100/22500) [54,900 – 36]
= (100/22500) × 54,864
= 243.84
Hence, standard deviation = σ = √243.84
= 15.61
∴ C.V for group B = (σ/ x̅) × 100
= (15.61/44.6) × 100
= 35
By comparing the C.V. of group A and group B,
C.V. of Group B > C.V. of Group A.
So, Group B is more variable.






