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Question: Let \[\bar X\] and \[MD\] be the mean and the mean deviation about \[\bar X\] of n observations, \[{...

Let Xˉ\bar X and MDMD be the mean and the mean deviation about Xˉ\bar X of n observations, xi,i=1,2,3,......n{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.Xˉ,MD\bar X,MD
B. Xˉ+5,MD\bar X + 5,MD
C. Xˉ,MD+5\bar X,MD + 5
D. Xˉ+5,MD+5\bar X + 5,MD + 5

Explanation

Solution

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 xi,i=1,2,3,......n{x_i},i = 1,2,3,......n is given by Xˉ=∑i=1nxin\bar X = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}
b) Mean deviation about the mean Xˉ\bar X is given by MD=∑i=1nxi−XˉMD = \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.Xˉ=∑i=1nxin\bar X = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}
Expanding the terms in RHS of the equation.
Xˉ=x1+x2+.......xnn……….(1)\bar X = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n} ………. (1)
Now we add 5 to each observation, therefore sum of observations become
⇒x1+5+x2+5+.......xn+5\Rightarrow {x_1} + 5 + {x_2} + 5 + .......{x_n} + 5
Since 5 is added n times we can write
⇒x1+x2+.......xn+5n\Rightarrow {x_1} + {x_2} + .......{x_n} + 5n
Now new mean be denoted by Xˉ′\bar X'
⇒Xˉ′=x1+x2+.......xn+5nn\Rightarrow \bar X' = \dfrac{{{x_1} + {x_2} + .......{x_n} + 5n}}{n}
Separate the term 5n from the fraction.
⇒Xˉ′=x1+x2+.......xnn+5nn\Rightarrow \bar X' = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n} + \dfrac{{5n}}{n}
Cancel out the same terms from numerator and denominator.
⇒Xˉ′=x1+x2+.......xnn+5\Rightarrow \bar X' = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n} + 5
Using equation (1) write Xˉ=x1+x2+.......xnn\bar X = \dfrac{{{x_1} + {x_2} + .......{x_n}}}{n}
⇒Xˉ′=Xˉ+5……………..(2)\Rightarrow \bar X' = \bar X + 5…………….. (2)
Now we know that mean deviation of n observations about the mean Xˉ\bar Xis given by MD=∑i=1nxi−XˉMD = \sum\limits_{i = 1}^n {{x_i} - \bar X} .
Expanding the term on RHS of the equation.
MD=(x1−Xˉ)+(x2−Xˉ)+.....+(xn−Xˉ)……….(3)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. Xˉ′=Xˉ+5\bar X' = \bar X + 5
Let us denote new mean deviation by MD′MD'
⇒MD′=(x1+5−Xˉ′)+(x2+5−Xˉ′)+.....+(xn+5−Xˉ′)\Rightarrow MD' = ({x_1} + 5 - \bar X') + ({x_2} + 5 - \bar X') + ..... + ({x_n} + 5 - \bar X')
Substitute the value of Xˉ′=Xˉ+5\bar X' = \bar X + 5from equation (2)
⇒MD′=(x1+5−(Xˉ+5))+(x2+5−(Xˉ+5))+.....+(xn+5−(Xˉ+5))\Rightarrow MD' = ({x_1} + 5 - (\bar X + 5)) + ({x_2} + 5 - (\bar X + 5)) + ..... + ({x_n} + 5 - (\bar X + 5))
Opening the brackets

⇒MD′=(x1+5−Xˉ−5)+(x2+5−Xˉ−5)+.....+(xn+5−Xˉ−5) ⇒MD′=(x1−Xˉ)+(x2−Xˉ)+.....+(xn−Xˉ)  \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=(x1−Xˉ)+(x2−Xˉ)+.....+(xn−Xˉ)MD = ({x_1} - \bar X) + ({x_2} - \bar X) + ..... + ({x_n} - \bar X) from equation (3)
⇒MD′=MD\Rightarrow MD' = MD

Therefore, new mean and new mean deviation about the new mean are Xˉ+5\bar X + 5 and MDMD 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.