Question

Find ∑(x-xi)2=100, no. of observations = 20, ∑xi=20.

Answer 1

We are given the following information: 
 $\sum (x - x_i)^2 = 100$ 
Number of observations (n) = 20 
$\sum x_i = 20$ 
We need to find the value of x.  

First, we can find the mean of the observations, which we'll denote as $\bar{x}$: 
$\bar{x} = \frac{\sum x_i}{n} = \frac{20}{20} = 1$ 
Now, we are given that $\sum (x - x_i)^2 = 100$. This looks like the formula for variance, but with 'x' instead of the mean. 
We can rewrite the given sum as:  

$\sum (x - x_i)^2 = \sum (x^2 - 2xx_i + x_i^2) = 100$ 
Expanding the summation: 
$\sum x^2 - 2x \sum x_i + \sum x_i^2 = 100$ 
Since 'x' is a constant with respect to the summation over 'i', we can write:  

\$n x^2 - 2x \sum x_i + \sum x_i^2 = 100$ 
We know that n = 20 and $\sum x_i = 20$. Substituting these values: 

$20x^2 - 2x(20) + \sum x_i^2 = 100$ 
$20x^2 - 40x + \sum x_i^2 = 100$ 
To proceed further, we need to find $\sum x_i^2$. We know that the variance (s²) is given by: 
 $s^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{\sum x_i^2}{n} - \bar{x}^2$ 
We can also express the given sum as: 
$\frac{\sum (x - x_i)^2}{n} = \frac{100}{20} = 5$ 
Let's rewrite the given sum in another way: 
 

$\frac{\sum (x - x_i)^2}{n} = \frac{\sum (x - \bar{x} + \bar{x} - x_i)^2}{n}$  

$= \frac{\sum [(x - \bar{x})^2 + 2(x - \bar{x})(\bar{x} - x_i) + (\bar{x} - x_i)^2]}{n}$ 

$= \frac{n(x - \bar{x})^2}{n} + \frac{2(x - \bar{x})\sum (\bar{x} - x_i)}{n} + \frac{\sum (\bar{x} - x_i)^2}{n}$ 

Since $\sum(\bar{x} - x_i) = 0$, the second term is zero. 
So, we have: 
$(x - \bar{x})^2 + s^2 = 5$ 
$(x - 1)^2 + s^2 = 5$ 
We also know that $s^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$. 
 

From the equation $20x^2 - 40x + \sum x_i^2 = 100$, we have: 
$\sum x_i^2 = 100 - 20x^2 + 40x$ 
$s^2 = \frac{100 - 20x^2 + 40x}{20} - 1^2 = 5 - x^2 + 2x - 1 = 4 - x^2 + 2x$ 
Substituting $s^2$ into $(x - 1)^2 + s^2 = 5$:  

$(x - 1)^2 + 4 - x^2 + 2x = 5$ 
$x^2 - 2x + 1 + 4 - x^2 + 2x = 5$ 
$5 = 5$ 
This equation is always true, which means that 'x' can be any real number. However, we can look for a simpler solution.  

If $x = \bar{x} = 1$, then $\sum (x - x_i)^2 = \sum (\bar{x} - x_i)^2 = n \cdot s^2$. 
So, $100 = 20 \cdot s^2$, which gives $s^2 = 5$. 
If $x = 1$, then $(x-1)^2 + s^2 = 0 + 5 = 5$, which satisfies the given condition. 
Therefore, the most likely value of x is 1.