Question: From the following data, find the standard deviation and coefficient of variation.
Class Interval | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency | 10 | 15 | 25 | 30 | 12 | 8 |
Solution:
Using actual mean,
[We have, interval between two classes is 10.
To get the frequency, subtract the give frequency of the respective
class from one step higher class.]
Arranging the data in a table,
X | f | m | fm | d|x-21.5| | d² | fd² |
20-30 | 10 | 25 | 250 | 24.3 | 590.49 | 5904.9 |
30-40 | 15 | 35 | 525 | 14.3 | 204.49 | 3067.35 |
40-50 | 25 | 45 | 1125 | 4.3 | 18.49 | 462.25 |
30-40 | 30 | 55 | 1650 | 5.7 | 32.49 | 974.7 |
40-50 | 12 | 65 | 780 | 15.7 | 246.49 |
2957.88 |
70-80 | 8 | 75 | 600 | 25.7 | 660.49 | 5283.92 |
N=100 |
|
\sumfm = 4930 |
\sumfd² =18651 |
Mean = \dfrac{\sum fm}{N}
= \dfrac{4930}{100}
= 49.3
Standard Deviation (\sigma) = \sqrt{ \dfrac{\sum fd²}{N} }
= \sqrt{ \dfrac{18651}{100} }
= \sqrt{186.51}
= 13.66
And,
Coefficient of Variation = \dfrac{\sigma}{mean} × 100%
= \dfrac{13.66}{49.3} × 100%
= 0.2770 × 100%
= 27.7%
Hence, the required standard deviation using actual mean of the given
data is 13.66 and the coefficient of variation is 27.7%.
Related Notes and Solutions:
Here is the website link to the notes of Statistics of Class 10.
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