Question

The mean of $5$ observation is $5$ and their variance is $12.4$. If three of the observations are $1$,$2$ and $6$; then their mean deviation from the mean of the data is:
A) $2.5$
B) $2.6$
C) $2.8$
D) $2.4$

Answer 1

Hint:Assume two numbers be $x$ and $y$. Mean can be given by $\dfrac{{\sum\limits_{i = 1}^n {{X_i}} }}{n}$ that means $\overline X $ that $\dfrac{{{x_1} + {x_2} + - - - - {x_n}}}{n}$.And variance is given by $\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{X_i} - \overline X } \right)}^2}} }}{n}$. Here $\overline X $ is given as $5$.Using these concepts we try to get answer.

Complete step-by-step answer:
Here the mean and variance of $5$ observations is given as $5$ and $12.4$ respectively. Out of $5$, three observations are given as $1$,$2$ and $6$.
Now let us assume $x$ and $y$ are the rest two observations.
Mean can be given by $\dfrac{{\sum\limits_{i = 1}^n {{X_i}} }}{n}$
Mean ($\overline X $)$ = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5}$
And mean is given as $5$
$
  5 = \dfrac{{1 + 2 + 6 + x + y}}{5} \\
  25 = 9 + x + y \\
  25 - 9 = x + y \\
 $
$x + y = 16…………….(1)$
So, we got one equation.
Now we are provided that variance of the observations is $12.4$.
Now we know,
Variance is given by $\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{X_i} - \overline X } \right)}^2}} }}{n}$.
Here we know $\overline X $ is the mean and equal to $5$.
So, variance
$
   = \left( {\dfrac{{{{\left( {{x_1} - 5} \right)}^2} + {{\left( {{x_2} - 5} \right)}^2} + {{\left( {{x_3} - 5} \right)}^2} + {{\left( {{x_4} - 5} \right)}^2} + {{\left( {{x_5} - 5} \right)}^2}}}{5}} \right) \\
  12.4 = \dfrac{{{{\left( {1 - 5} \right)}^2} + {{\left( {2 - 5} \right)}^2} + {{\left( {6 - 5} \right)}^2} + {{\left( {x - 5} \right)}^2} + {{\left( {y - 5} \right)}^2}}}{5} \\
  12.4 \times 5 = 16 + 9 + 1 + {\left( {x - 5} \right)^2} + {\left( {y - 5} \right)^2} \\
  62 = 26 + {\left( {x - 5} \right)^2} + {\left( {y - 5} \right)^2} \\
 $
And we know $y = 16 - x$ from equation (1). So,
$
  62 = 26 + {\left( {x - 5} \right)^2} + {\left( {16 - x - 5} \right)^2} \\
  36 = {x^2} + 25 - 10x + 121 + {x^2} - 22x \\
  2{x^2} - 32x + 110 = 0 \\
  {x^2} - 16x + 55 = 0 \\
  {x^2} - 11x - 5x + 55 = 0 \\
  x\left( {x - 11} \right) - 5\left( {x - 11} \right) = 0 \\
  \left( {x - 11} \right)\left( {x - 5} \right) = 0 \\
  x = 11,5 \\
 $
We get $x = 11$ and $x = 5$
Now, if we take $x = 11$, we get $y = 16 - x = 16 - 11 = 5$
If we take $x = 5$, we get $y = 16 - x = 16 - 5 = 11$
So, in each case we get $5,11$ as the other two numbers in the observations.
Now, we need to find mean deviation.
Mean deviation$ = \dfrac{{\sum\limits_{i = 1}^n {|{X_i} - \overline X |} }}{5}$
$
   = \dfrac{{|1 - 5| + |2 - 5| + |6 - 5| + |11 - 5| + |5 - 5|}}{5} \\
   = \dfrac{{4 + 3 + 1 + 6 + 0}}{5} = \dfrac{{14}}{5} = 2.8 \\
 $

So, the correct answer is “Option C”.

Note:Whenever we talk about the mean deviation, it is the value deviated from the mean. And remember that deviation always has a positive value. So we use modulus in the formula.Mean deviation$ = \dfrac{{\sum\limits_{i = 1}^n {|{X_i} - \overline X |} }}{5}$, where $\overline X $ is the mean of observation.