Question

If $\overline{X}$ is the mean of ${{x}_{1}},{{x}_{2}},{{x}_{3}},...,{{x}_{n}}$ then algebraic sum of deviations about mean $\overline{X}$ is \[\]
A.$0$\[\]
B.$\dfrac{\overline{X}}{n}$\[\]
C.$n\overline{X}$\[\]
D.None of these \[\]

Answer 1

Hint: We recall the definition sample mean $\overline{X}=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}$ and deviations of any data value also called an observation from the mean $\overline{X}$ as $d\left( {{x}_{i}} \right)={{x}_{i}}-\overline{X}$. We add the deviations of all the observations and try to simplify using the definition of mean .\[\]

Complete step by step answer:
We know that mean or sample mean is one of the measure of central tendency also known as the average, expectation of the data sample . It is denoted by $\overline{X}$. If there are $n$ number of data values with equal weights in the samples say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ then the mean is calculated by first finding the sum of data values and then by dividing the sum by $n.$ So the sample is given by
\[\begin{align}
  & \overline{X}=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}=n\overline{X}......\left( 1 \right) \\
\end{align}\]
Deviation is a measure of difference between the observed value of a variable and expected value. The deviation of any data value ${{x}_{i}}$ is the difference of that data from the mean is called its deviation. We define deviation as
\[d\left( {{x}_{i}} \right)={{x}_{i}}-\overline{X}\]
Let us find the algebraic sum of all deviations. We have;
\[\begin{align}
  & \Rightarrow d\left( {{x}_{1}} \right)+d\left( {{x}_{2}} \right)+...+d\left( {{x}_{n}} \right) \\
 & \Rightarrow {{x}_{1}}-\overline{X}+{{x}_{2}}-\overline{X}+...+{{x}_{n}}-\overline{X} \\
\end{align}\]
Let us separate the data values from mean and have;
\[\Rightarrow {{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}-\left( \overline{X}+\overline{X}+...\text{ }n\text{ times} \right)\]
We already have ${{x}_{1}}+{{x}_{2}}...+{{x}_{n}}=n\overline{X}$ and we multiply $n$ times $\overline{X}$ we will get the product$n\overline{X}$We use these results in the above step to have;
\[\Rightarrow n\overline{X}-n\overline{X}=0\]

So, the correct answer is “Option A”.

Note: We note that since algebraic sum about mean is zero, it cannot tell us how the data is dispersed from the mean, for that we need other type of deviations like the absolute mean deviation is given as $D=\left| {{x}_{i}}-\overline{X} \right|$, the standard deviation $\sigma =\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{X} \right)}^{2}}}}{n-1}$. The algebraic sum of squared deviations is called variance ${{\sigma }^{2}}={{\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-\overline{X} \right)}}^{2}}$.