Question

If the algebraic sum of deviations of 20 observations from 30 is 20, then the mean of the observation is
A. 30
B. $30.1$
C. 29
D. 31

Answer 1

Hint: We first find the general form of deviations of n observations and the form mean for those n observations. We find the formula for the mean using the values of the number of observations and the deviation value. We place the values and get the value of the mean.

Complete step-by-step solution:
Let a set of n observations be ${{x}_{i}},i=1 \text{to} n$. The mean of the set of the observations will be \[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\].
We are taking deviations of those observations from the constant m. The deviations will be ${{x}_{i}}-m$. The sum of the deviations will be \[{{x}_{D}}=\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-m \right)}=\sum\limits_{i=1}^{n}{{{x}_{i}}}-mn=n\overline{x}-mn=n\left( \overline{x}-m \right)\].
For our given problem it is given that the algebraic sum of deviations of 20 observations from 30 is 20.
The replacement of the variables will be for $n=20,m=30,{{x}_{D}}=20$.
We need to find the mean of the observation which is equal to \[\overline{x}\].
Putting the values in the equation of \[{{x}_{D}}=n\left( \overline{x}-m \right)\], we get \[20=20\left( \overline{x}-30 \right)\].
We solve the equation by dividing both sides of the equation with 20 and get
\[\begin{align}
  & 20=20\left( \overline{x}-30 \right) \\
 & \Rightarrow \overline{x}-30=\dfrac{20}{20}=1 \\
\end{align}\]
Now we find the value of \[\overline{x}\] by addition and get \[\overline{x}=1+30=31\].
The correct option is D.

Note: We need to remember that the value of m in the equation of \[\sum\limits_{i=1}^{n}{\left( {{x}_{i}}-m \right)}=\sum\limits_{i=1}^{n}{{{x}_{i}}}-mn\] was constant and the summation wouldn’t have worked. That’s why we multiplied the number of iterations of n with m to find the summation.