# Statistics

The measures of the data that shows the range or spread of the data from the central value is called measure of the variability and dispersion

Range, mean deviation, quartile deviation and standard deviation are the four measures of dispersion.

Range

The difference between the highest and lowest value of any set of data is called range of the data.

R = L- S

Coefficient of range=$\frac{{{\rm{\: L}} - {\rm{S}}}}{{{\rm{L}} + {\rm{S}}}}$

For example, let us consider the following data set:

23, 65,82,59,55,25,85,56

Largest value (L) =85

Smallest value (S) =23

R = L- S85 -23 = 62

Mean

A mean is defined as the average of the numbers. The data may be individual, discrete and continuous .The method of calculating the means depend upon the nature of data. It is denoted byx̅

For individual data:x̅ =$\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + {\rm{\: }}{{\rm{x}}_4} \ldots \ldots {{\rm{x}}_{\rm{n}}}}}{{\rm{N}}}$=$\frac{{\Sigma {\rm{x}}}}{{\rm{N}}}{\rm{\: }}$

For discrete data:x̅ =$\frac{{{{\rm{f}}_1}{{\rm{x}}_1} + {{\rm{f}}_2}{{\rm{x}}_2} + {{\rm{f}}_3}{{\rm{x}}_3} + {\rm{\: }}{{\rm{x}}_4} \ldots \ldots {{\rm{f}}_{\rm{n}}}{{\rm{x}}_{\rm{n}}}}}{{{{\rm{f}}_1} + {\rm{\: }}{{\rm{f}}_2} + {\rm{\: }}{{\rm{f}}_3} \ldots \ldots \ldots \ldots ..{{\rm{f}}_{\rm{n}}}}}$$= {\rm{\: }}\frac{{\Sigma {\rm{fx}}}}{{\rm{N}}} For continuous data:\frac{{{{\rm{f}}_1}{{\rm{x}}_1} + {{\rm{f}}_2}{{\rm{x}}_2} + {{\rm{f}}_3}{{\rm{x}}_3} + {\rm{\: }}{{\rm{x}}_4} \ldots \ldots {{\rm{f}}_{\rm{n}}}{{\rm{x}}_{\rm{n}}}}}{{{{\rm{f}}_1} + {\rm{\: }}{{\rm{f}}_2} + {\rm{\: }}{{\rm{f}}_3} \ldots \ldots \ldots \ldots ..{{\rm{f}}_{\rm{n}}}}}=\frac{{\Sigma {\rm{fx}}}}{{\rm{N}}} Mean deviation The mean deviation is the mean of the absolute deviationsof a set of data about the mean. For a sample size N, the mean deviation is defined by M.D =\frac{1}{{\rm{N}}}\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} \left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right| Where x̅ is the mean of the distribution The mean deviation for a discrete and continuous distribution defined by M.D =\frac{1}{{\rm{N}}}\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} {\rm{f}}\left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x\: }}} \right| Coefficient of mean deviation =\frac{{{\rm{M}}.{\rm{D}}}}{{{\rm{\bar X}}}}{\rm{\: }} Standard deviation The Standard Deviation is a measure of how spreads out numbers are. It is denoted by σ. it is the square root of the Variance. For individual series σ =\sqrt {\frac{{\Sigma {{\rm{d}}^2}}}{{\rm{N}}}} {\rm{\: }}where d = \left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x\: }}} \right|where I = 1…N For discrete and continuous series σ =\sqrt {\frac{{\Sigma {\rm{f}}{{\rm{d}}^2}}}{{\rm{N}}}} {\rm{\: }}where d = \left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x\: }}} \right|where I = 1…N for discrete and d = \left| {{\rm{m}} - {\rm{\bar x\: }}} \right|m = mid value Coefficient of variance= \frac{\sigma }{{{\rm{\bar x}}}}{\rm{\: }} Median Median is the mid –value of the data. For the individual and discrete data Median = value of \frac{{{\rm{N}} + 1}}{2} item if the number of the terms is odd = average value of\frac{{{\rm{\: N}}}}{2}$${\rm{and\: }}\frac{{{\rm{N}} + 2}}{2}$item if the number of the terms is even

For continuous data

Median = L + $\frac{{\frac{{\rm{N}}}{2} - {\rm{c}}.{\rm{f}}}}{{\rm{f}}}$ *i

N= total number of items

L = lower limit of median class

C.F = C.F of the preceding the median class

F= frequency of median class

I = class interval

Example 1

Find the mean deviation and median from the median and its coefficient of the data from given below.

 x f 4 6 6 4 8 5 10 3 12 2 14 1 16 4

a.

 x f c.f |D| f|D| Md 4 6 6 4 24 Value of $\frac{{{\rm{N}} + 1{\rm{\: }}}}{2}{\rm{th\: item}}$ = 13th term =8 6 4 10 2 8 8 5 15 0 0 10 3 18 2 6 12 2 20 4 8 14 1 21 6 6 16 4 25 8 32 N=25 84

Mean deviation (M.D) =$\frac{{\Sigma {\rm{f}}\left| {\rm{D}} \right|}}{{\rm{N}}}{\rm{\: }}$= 3.36

Coefficient of mean deviation from the median is given by coefficient of M.D = $\frac{{{\rm{M}}.{\rm{D}}\left( {{{\rm{M}}_{\rm{d}}}} \right)}}{{{{\rm{M}}_{\rm{d}}}}}$= $\frac{{3.36}}{8}{\rm{\: }} = 0.42$

Example 2

Find the standard deviation and coefficient of standard deviation from the following data.

 x mean(X̅) d = X – X̅ ${{\rm{d}}^2}$ 14 66.28 -52.3 2735.29 46 -20.3 412.09 58 -8.3 68.89 76 9.7 94.09 82 15.7 246.49 90 23.7 561.69 98 31.7 1004.89 Σ X = 464 5123.43

Standard deviation =$\sqrt {\frac{{\Sigma {{\rm{d}}^2}}}{{\rm{N}}}}$=$\sqrt {\frac{{5123.43}}{7}}$= 27.05

Coefficient of variation= $\frac{\sigma }{{{\rm{\: \bar X}}}}$* 100 =$\frac{{27.05}}{{66.28}}$ *100 = 40

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