Question

Let \[\bar X\] and \[MD\] be the mean and the mean deviation about \[\bar X\] of n observations, \[{x_i},i = 1,2,3,......n\]. If each of the observations is increased by 5, then the new mean and MD about new mean respectively are:
A.\[\bar X,MD\]
B. \[\bar X + 5,MD\]
C. \[\bar X,MD + 5\]
D. \[\bar X + 5,MD + 5\]

Answer 1

Hint:We first write mean of n observations using the formula and then find new mean, after each observation is added with 5 in it. Then we write mean deviation of n observations with original mean and then for new mean deviation add 5 to each term and write new mean in place of original mean.

Formula used:
a) Mean of n observations \[{x_i},i = 1,2,3,......n\] is given by \[\bar X = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\]
b) Mean deviation about the mean \[\bar X\] is given by \[MD = \sum\limits_{i = 1}^n {{x_i} - \bar X} \]

Complete step-by-step answer:
We have a number of observations as n.
We can write the original mean as sum of observations divided by number of observations i.e.\[\bar X = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\]
Expanding the terms in RHS of the equation.
\[\bar X = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n} ………. (1)\]
Now we add 5 to each observation, therefore sum of observations become
\[ \Rightarrow {x_1} + 5 + {x_2} + 5 + .......{x_n} + 5\]
Since 5 is added n times we can write
\[ \Rightarrow {x_1} + {x_2} + .......{x_n} + 5n\]
Now new mean be denoted by \[\bar X'\]
\[ \Rightarrow \bar X' = \dfrac{{{x_1} + {x_2} + .......{x_n} + 5n}}{n}\]
Separate the term 5n from the fraction.
\[ \Rightarrow \bar X' = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n} + \dfrac{{5n}}{n}\]
Cancel out the same terms from numerator and denominator.
\[ \Rightarrow \bar X' = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n} + 5\]
Using equation (1) write \[\bar X = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n}\]
\[ \Rightarrow \bar X' = \bar X + 5…………….. (2)\]
Now we know that mean deviation of n observations about the mean \[\bar X\]is given by \[MD = \sum\limits_{i = 1}^n {{x_i} - \bar X} \].
Expanding the term on RHS of the equation.
\[MD = ({x_1} - \bar X) + ({x_2} - \bar X) + ..... + ({x_n} - \bar X) ………. (3)\]
Now we add 5 to each observation and find the mean deviation about the new mean obtained after adding 5 to each observation, i.e. \[\bar X' = \bar X + 5\]
Let us denote new mean deviation by \[MD'\]
\[ \Rightarrow MD' = ({x_1} + 5 - \bar X') + ({x_2} + 5 - \bar X') + ..... + ({x_n} + 5 - \bar X')\]
Substitute the value of \[\bar X' = \bar X + 5\]from equation (2)
\[ \Rightarrow MD' = ({x_1} + 5 - (\bar X + 5)) + ({x_2} + 5 - (\bar X + 5)) + ..... + ({x_n} + 5 - (\bar X + 5))\]
Opening the brackets
\[
   \Rightarrow MD' = ({x_1} + 5 - \bar X - 5) + ({x_2} + 5 - \bar X - 5) + ..... + ({x_n} + 5 - \bar X - 5) \\
   \Rightarrow MD' = ({x_1} - \bar X) + ({x_2} - \bar X) + ..... + ({x_n} - \bar X) \\
 \]
Substituting the value of \[MD = ({x_1} - \bar X) + ({x_2} - \bar X) + ..... + ({x_n} - \bar X)\] from equation (3)
\[ \Rightarrow MD' = MD\]

Therefore, new mean and new mean deviation about the new mean are \[\bar X + 5\] and \[MD\] respectively.

So, the correct answer is “Option B”.

Note:Students are likely to make the mistake of adding the number to the number of observations i.e. n also, which is wrong because we are not adding the number of observations, we are adding values in the observation.