Question

If the sum of squares of the deviations from the mean of the data \( x_i \) (\( i = 1, 2, \dots, n \)) is \( n\overline{x}^2 \), where \( \overline{x} \) is the mean of \( x_i \)'s, then the sum of squares of \( x_i \)'s is

• A. \( 4n\overline{x}^2 \)
• B. \( 3n\overline{x}^2 \)
• C. \( 12n\overline{x}^2 \)
• D. \( 2n\overline{x}^2 \)

Hint:

Recall the definition of the sum of squares of deviations from the mean: \( \sum (x_i - \overline{x})^2 \). Expand this expression and use the definition of the mean \( \overline{x} = \frac{1}{n} \sum x_i \) to simplify. Relate the given information \( \sum (x_i - \overline{x})^2 = n\overline{x}^2 \) to the required sum of squares \( \sum x_i^2 \).

Answer 1

The Correct Option is D

The sum of squares of the deviations from the mean is given by \( \sum_{i=1}^{n} (x_i - \overline{x})^2 \).
We are given that \( \sum_{i=1}^{n} (x_i - \overline{x})^2 = n\overline{x}^2 \).
Expanding the term \( (x_i - \overline{x})^2 \), we get \( x_i^2 - 2x_i\overline{x} + \overline{x}^2 \).
So, \( \sum_{i=1}^{n} (x_i^2 - 2x_i\overline{x} + \overline{x}^2) = n\overline{x}^2 \).
Using the linearity of summation: $$ \sum_{i=1}^{n} x_i^2 - \sum_{i=1}^{n} 2x_i\overline{x} + \sum_{i=1}^{n} \overline{x}^2 = n\overline{x}^2 $$ We know that \( \sum_{i=1}^{n} 2x_i\overline{x} = 2\overline{x} \sum_{i=1}^{n} x_i = 2\overline{x} (n\overline{x}) = 2n\overline{x}^2 \).
Also, \( \sum_{i=1}^{n} \overline{x}^2 = n\overline{x}^2 \) (since \( \overline{x} \) is a constant with respect to the summation index \( i \)).
Substituting these into the equation: $$ \sum_{i=1}^{n} x_i^2 - 2n\overline{x}^2 + n\overline{x}^2 = n\overline{x}^2 $$ $$ \sum_{i=1}^{n} x_i^2 - n\overline{x}^2 = n\overline{x}^2 $$ $$ \sum_{i=1}^{n} x_i^2 = n\overline{x}^2 + n\overline{x}^2 $$ $$ \sum_{i=1}^{n} x_i^2 = 2n\overline{x}^2 $$ The sum of squares of \( x_i \)'s is \( 2n\overline{x}^2 \).