Sheppard Correction and Corrected Coefficient of Variation
Sheppard Corrections
In grouped data the different observations are put into the same class. In the calculation of variation or standard deviation for grouped data, the frequency $$f$$ is multiplied by $$X$$, which is the midpoint of the respective class. Thus it is assumed that all the observations in a class are centered at $$X$$. But this is not true, because the observations are spread in the said class. This assumption introduces some error in the calculation of $${S^2}$$ and $$S$$. The value of $${S^2}$$ and $$S$$ can be corrected to some extent by applying Sheppard correction. Thus
\[{S^2}\left( {{\text{corrected}}} \right) = {\text{ }}{S^2} – \frac{{{h^2}}}{{12}}\]
\[S\left( {{\text{corrected}}} \right) = {\text{ }}\sqrt {{S^2} – \frac{{{h^2}}}{{12}}} \]
Here $$h$$ is the uniform class interval.
This correction is applied in grouped data which has almost equal tails at the start and at the end of the data. If a data has a longer tail on any side, this correction is not applied. If the size of the class interval $$h$$ is not the same in all classes, the correction is not applicable.
Corrected Coefficient of Variation
When the corrected standard deviation is used in the calculation of the coefficient of variation, we get what is called the corrected coefficient of variation. Thus the corrected coefficient of variation
\[Corrected\,Coefficient\,of\,Variation = {\text{ }}\frac{{S\left( {{\text{corrected}}} \right)}}{{\overline X }} \times 100\]
Example:
Calculate the Sheppard correction and corrected coefficient of variation from the following distribution of marks by using all the methods.
Marks 
No. of Students

$$1 – 3$$

$$40$$

$$3 – 5$$

$$30$$

$$5 – 7$$

$$20$$

$$7 – 9$$

$$10$$

Solution:
Marks

$$f$$

$$X$$

$$fX$$

$$U = \left( {X – 2} \right)/2$$

$$fU$$

$$f{U^2}$$

$$1 – 3$$

$$40$$

$$2$$

$$80$$

$$ – 2$$

$$ – 80$$

$$160$$

$$3 – 5$$

$$30$$

$$4$$

$$120$$

$$ – 1$$

$$ – 30$$

$$30$$

$$5 – 7$$

$$20$$

$$6$$

$$120$$

$$0$$

$$0$$

$$0$$

$$7 – 9$$

$$10$$

$$8$$

$$80$$

$$1$$

$$10$$

$$10$$

Total

$$100$$


$$400$$


$$ – 100$$

$$200$$

$$\overline X = \frac{{\sum fX}}{{\sum f}} = \frac{{400}}{{100}} = 4$$
$${S^2} = \left[ {\frac{{\sum f{U^2}}}{{\sum f}} – {{\left( {\frac{{\sum fU}}{{\sum f}}} \right)}^2}} \right] \times h = \left[ {\frac{{200}}{{100}} – {{\left( {\frac{{ – 100}}{{100}}} \right)}^2}} \right] \times 2$$
$${S^2} = \left[ {2 – 1} \right] \times 2 = 1 \times 2 = 2$$
$${S^2}\left( {{\text{corrected}}} \right) = {\text{ }}{S^2} – \frac{{{h^2}}}{{12}} = 2 – \frac{4}{{12}} = \frac{5}{3} = 1.67$$
$$S\left( {{\text{corrected}}} \right) = {\text{ }}\sqrt {{S^2} – \frac{{{h^2}}}{{12}}} = \sqrt {1.67} = 1.29$$
$$Corrected\,Coefficient\,of\,Variation{\text{ = }}\frac{{S\left( {{\text{corrected}}} \right)}}{{\overline X }} \times 100$$
$${\text{ = }}\frac{{1.29}}{4} \times 100 = 32.25$$
Sanjay madhav binnar
February 4 @ 3:19 pm
Incorrect solution