Understanding the Sum of Deviations from the Mean

The question asks for the value of the sum of the deviations of each observation from the mean for a set of data points \(\rm X_1, X_2, X_3, \ldots, X_n\). The mean of these observations is denoted by \(\rm \bar X\). The expression we need to evaluate is \(\rm \frac{\Sigma ^n}{i=1}(X_i-\bar X)\).

This expression represents the sum of deviations from the mean. A deviation for a single observation \(\rm X_i\) is the difference between that observation and the mean, \(\rm (X_i - \bar X)\). We are asked to find the sum of all such deviations for all \(n\) observations.

Calculating the Sum of Deviations

Let's expand the summation:

$$ \sum_{i=1}^n (X_i - \bar X) = (X_1 - \bar X) + (X_2 - \bar X) + \ldots + (X_n - \bar X) $$

We can rearrange the terms by grouping the \(\rm X_i\) values together and the \(\rm \bar X\) values together:

$$ \sum_{i=1}^n (X_i - \bar X) = (X_1 + X_2 + \ldots + X_n) - (\bar X + \bar X + \ldots + \bar X) $$

The sum of all observations \(\rm (X_1 + X_2 + \ldots + X_n)\) is written in sigma notation as \(\rm \frac{\Sigma ^n}{i=1} X_i\). The term \(\rm (\bar X + \bar X + \ldots + \bar X)\) is the mean \(\rm \bar X\) added to itself \(n\) times, which is simply \(\rm n\bar X\).

So, the expression becomes:

$$ \sum_{i=1}^n (X_i - \bar X) = \left(\sum_{i=1}^n X_i\right) - n\bar X $$

Relating the Sum of Observations to the Mean

Now, let's recall the definition of the arithmetic mean \(\rm \bar X\) for a set of \(n\) observations:

$$ \bar X = \frac{\sum_{i=1}^n X_i}{n} $$

If we multiply both sides of this equation by \(n\), we get a fundamental relationship:

$$ n\bar X = \sum_{i=1}^n X_i $$

This shows that the total sum of the observations \(\rm (\Sigma X_i)\) is equal to \(n\) times the mean \(\rm (\bar X)\). This relationship is a key property of the mean.

Finding the Final Sum

Now we can substitute the relationship \(\rm \sum_{i=1}^n X_i = n\bar X\) back into our expression for the sum of deviations:

$$ \sum_{i=1}^n (X_i - \bar X) = \left(\sum_{i=1}^n X_i\right) - n\bar X $$

Substitute \(\rm \sum_{i=1}^n X_i\) with \(\rm n\bar X\):

$$ \sum_{i=1}^n (X_i - \bar X) = n\bar X - n\bar X $$

$$ \sum_{i=1}^n (X_i - \bar X) = 0 $$

Therefore, the sum of deviations from the mean for any set of observations is always zero. This is an important statistics formula and one of the essential properties of mean.

The sigma notation \(\rm \frac{\Sigma ^n}{i=1}\) indicates the sum over all observations from \(i=1\) to \(n\). This calculation confirms that the sum of deviations for all the given observations, \(\rm X_1, \ldots, X_n\), from their mean, \(\rm \bar X\), is indeed zero.