Question

Is the sum of deviations always zero?

Answer 1

Hint: First understand the meaning of the term ‘deviation’. Assume that there are n observations given as ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, …., ${{x}_{n}}$. Find the mean of these n observations by using the formula $\overline{x}=\dfrac{\left( \sum\limits_{i=1}^{n}{{{x}_{i}}} \right)}{n}$, where $\overline{x}$ is the mean and n is the number of observations. Now, use the formula for the sum of deviations (D) given as $D=\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-\overline{x} \right)}$ and use the relation of mean to check if we get 0 or not.

Complete step-by-step solution:
Here we have been asked to check if the sum of deviations is always zero or not. First let us understand the meaning of the term ‘deviation’.
In statistical mathematics, the deviation is a measure of the difference between the observed value in a data and the mean of the data. Mathematically, the sum of deviations (D) is given as $D=\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-\overline{x} \right)}$, where $\overline{x}$ is the mean. The deviation from a particular value in the data set can be negative or positive depending on the data values which are smaller or greater than the mean respectively.
Now, let us assume that we have n observations in a data set given as ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, …., ${{x}_{n}}$. Using the formula for the mean given as $\overline{x}=\dfrac{\left( \sum\limits_{i=1}^{n}{{{x}_{i}}} \right)}{n}$ we get,
$\begin{align}
  & \Rightarrow \overline{x}=\dfrac{\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+.....+{{x}_{n}} \right)}{n} \\
 & \Rightarrow n\overline{x}=\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+.....+{{x}_{n}} \right)........\left( i \right) \\
\end{align}$
Using the formula for the sum of deviations given as $D=\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-\overline{x} \right)}$ we get,
$\begin{align}
  & \Rightarrow D=\left( {{x}_{1}}-\overline{x} \right)+\left( {{x}_{2}}-\overline{x} \right)+\left( {{x}_{3}}-\overline{x} \right)+....+\left( {{x}_{n}}-\overline{x} \right) \\
 & \Rightarrow D=\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}} \right)-\left( \overline{x}+\overline{x}+\overline{x}+....+\overline{x} \right) \\
\end{align}$
In the above relation the term $\overline{x}$ will appear n times because $i$ is from 1 to n, so we get,
$\Rightarrow D=\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}} \right)-\left( n\overline{x} \right)$
Using relation (i) we can simplify the above expression as,
$\begin{align}
  & \Rightarrow D=\left( n\overline{x} \right)-\left( n\overline{x} \right) \\
 & \therefore D=0 \\
\end{align}$
Hence, the sum of deviations is always zero.

Note: Note that there are different types of deviations which are used to measure the dispersions in statistics like: - standard deviation, average absolute deviation, median absolute deviation etc. These deviations have different formulas for their measurements and their values need not be 0. In the above solution we have simply considered the basic definition of the term ‘deviation’ and the approach to calculate it.