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The average of n numbers x1, x2.....xn is $$\overline x $$ .Then the value of $$\sum\limits_{i\, = \,1}^n {} \left( {{x_i} - \overline x } \right)$$ is equal to -
Answer & Solution
Correct Answer:
Option
B
According to the question,
Average of 'n' number's x1, x2.....xn is $$\overline x $$
Sum of n numbers = n$${\overline x }$$
∴ $$\sum\limits_{i\, = \,1}^n {} \left( {{x_1} - \overline x } \right)$$
Put i = 1, 2, 3.....n then
{x1 + x2 + x3 + .....(xn - n$${\overline x }$$)}
As we know that
x1 + x2 + x3+.....+ xnx = n$${\overline x }$$
= (n$${\overline x }$$ - n$${\overline x }$$)
= 0
Average of 'n' number's x1, x2.....xn is $$\overline x $$
Sum of n numbers = n$${\overline x }$$
∴ $$\sum\limits_{i\, = \,1}^n {} \left( {{x_1} - \overline x } \right)$$
Put i = 1, 2, 3.....n then
{x1 + x2 + x3 + .....(xn - n$${\overline x }$$)}
As we know that
x1 + x2 + x3+.....+ xnx = n$${\overline x }$$
= (n$${\overline x }$$ - n$${\overline x }$$)
= 0
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