$\left( b \right)\,\,\bar x = \frac{{{x_1} + {x_2} + {x_3} + … + {x_n}}}{n}$

${\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}$