Question

in usual notation £d=25,£=272 n= 100 and assumed mean is 4. find coefficients of varition​

Answer 1

SOLUTION

GIVEN

$\sf{In \: usual \: notation \: \sum d = 25, \sum {d}^{2} = 272 ,n = 100,assumed \: mean = 4 }$

TO DETERMINE

The coefficient of variation

EVALUATION

Here it is given that

$\sf{In \: usual \: notation \: \sum d = 25, \sum {d}^{2} = 272 ,n = 100,assumed \: mean = 4 }$

Now

Mean

$\sf{ = \bar{x}}$

$\displaystyle \sf{ = Assumed \: mean + \frac{ \sum d}{n} }$

$\displaystyle \sf{ = 4 + \frac{25}{100} }$

$\displaystyle \sf{ =4 + 0.25 }$

$\displaystyle \sf{ =4.25 }$

Now

Variance

$\displaystyle \sf{ = \frac{ \sum {d}^{2} }{n} - { \bigg( \frac{ \sum d}{n} \bigg)}^{2} }$

$\displaystyle \sf{ = \frac{272 }{100} - { \bigg( \frac{ 25}{100} \bigg)}^{2} }$

$\displaystyle \sf{ = 2.72 - \frac{1}{16} }$

$\sf{ = 2.6575}$

Now standard deviation

$\sf{ = \sqrt{Variance} }$

$\sf{ = \sqrt{2.6575} }$

$\sf{= 1.63}$

Hence the required coefficient of variation

$\displaystyle \sf{ = \frac{ \sigma}{ \bar{x}} \times 100\% }$

$\displaystyle \sf{ = \frac{1.63}{4.25} \times 100\% }$

$= \sf{38.35\%}$

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