माना n प्रेक्षणों x_{1}, x_{2}, ldots, x_{ n | vedclass

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Question

$\left( b \right)\,\,\bar x = \frac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}$

${\sigma ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \

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कथन $I$ तथा कथन $II$ दोनों सत्य हैं।

Vedclass · Answer

$\left( b \right)\,\,\bar x = \frac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}$

${\sigma ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} $

Mean of ${d_1},{d_2},{d_3},......,{d_n}$

$ = \frac{{{d_1} + {d_2} + {d_3} + ...... + {d_n}}}{n}$

$ = \frac{{\left( { - {x_1} - a} \right) + \left( { - {x_2} - a} \right) + \left( { - {x_3} - a} \right) + ..... + \left( { - {x_n} - a} \right)}}{n}$

$ =  - \left[ {\frac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}} \right] - \frac{{na}}{n}$

$ =  - \bar x - a$

Since, ${d_i} =  - {x_i} - a$ and we multiply or subtract each observation by any number the mode remains the same. Hence mode of $ - {x_i} - a$ i.e. ${d_i}$ and ${x_i}$ are same.

Now variance of ${d_1},{d_2},......,{d_n}$

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{d_i} - \left( { - \bar x - a} \right)} \right]}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ { - {x_i} - a + \bar x - a} \right]}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( { - {x_i} + \bar x} \right)}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\bar x - {x_i}} \right)}^2}}  = {\sigma ^2}$

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