Let the observations xi (1 [ ] i [ ] 10) satisfy the equations, [_i = 1^10 ( x_1 - 5 ) ] =…

Question

Let the observations xi (1 $$ \le $$ i $$ \le $$ 10) satisfy the

equations, $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10 and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40.

If $$\mu $$ and $$\lambda $$ are the mean and the variance of the

observations, x1 -- 3, x2 -- 3, ...., x10 -- 3, then

the ordered pair ($$\mu $$, $$\lambda $$) is equal to :

Accepted Answer

$$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10

$$ \Rightarrow $$ x1 + x2 + .... + x10 = 60 ....(1)

$$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40

$$ \Rightarrow $$ ($x_1^2 + x_2^2 + ... + x_{10}^2$) + 25 $$ \times $$ 10 -
10( x1 + x2 + .... + x10) = 40

$$ \Rightarrow $$ $x_1^2 + x_2^2 + ... + x_{10}^2$ = 390 .....(2)

From question,

$$\mu $$ = $${{\left( {{x_1} - 3} \right) + \left( {{x_2} - 3} \right) + ... + \left( {{x_{10}} - 3} \right)} \over {10}}$$

= $${{60 - 3 \times 10} \over {10}}$$ = 3

And $$\lambda $$ = variance = $${{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} \right)}^2}} } \over {10}}$$ - $$\mu $$2

= $${{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} \right) + 90 - 6\left( {\sum {{x_i}} } \right)} \over {10}}$$ - 9

= $${{390 + 90 - 360} \over {10}}$$ - 9

= 12 - 9 = 3

Solutions · 1