Question

The algebraic sum of deviation of set of \[n\] observation from their means is
A.0
B.\[\dfrac{{n\left( {n + 1} \right)}}{2}\]
C.\[\dfrac{{n\left( {n - 1} \right)}}{2}\]
D.\[\dfrac{{n + 1}}{2}\]

Answer 1

Hint: First we will use formula to calculate mean by adding up all the numbers and then divide by how many numbers there are. Then we will compute the value algebraic sum of deviations from the obtained mean and simplify to find the required value.

Complete step-by-step answer:
We are given that set of \[n\] observations from their means.
Let us assume that the values are \[{a_1}\], \[{a_2}\], \[{a_3}\],…, \[{a_n}\].
We know that the formula to calculate the mean is by adding up all the numbers and then dividing them by total numbers.
First, we will add the marks obtained by students, we get
\[ \Rightarrow {a_1} + {a_2} + {a_3} + ... + {a_n}\]
Dividing the above value by \[n\] to find the mean, we get
\[ \Rightarrow {\text{Mean}} = \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}\]
Computing the value of algebraic sum of deviations from the above mean, we get
\[ \Rightarrow \left( {{a_1} - \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}} \right) + \left( {{a_2} - \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}} \right) + ... + \left( {{a_n} - \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}} \right)\]
Rearranging the terms of the above equation, we get
\[ \Rightarrow {a_1} + {a_2} + {a_3} + ... + {a_n} - \left( {\dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n} + \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n} + ... + \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}} \right)\]
Since we know that the terms in the bracket of the above equation are \[n\] terms, we have
\[
   \Rightarrow {a_1} + {a_2} + ... + {a_n} - n\left( {\dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}} \right) \\
   \Rightarrow {a_1} + {a_2} + ... + {a_n} - \left( {{a_1} + {a_2} + ... + {a_n}} \right) \\
   \Rightarrow {a_1} + {a_2} + ... + {a_n} - {a_1} - {a_2} - ... - {a_n} \\
   \Rightarrow 0 \\
 \]
Therefore, the required value is 0.
Hence, option A is correct.

Note: We need to know that if we compute the mean of the deviations by summing the deviations and dividing by the sample size we run into a problem. The sum of the deviations from the mean is zero. This will always be the case as it is a property of the sample mean, i.e., the sum of the deviations below the mean will always equal the sum of the deviations above the mean. So this was a direct question, but we have added steps to it if asked for more marks.