Question

The mean square deviation of set of $n$observations ${x_1},{x_2}....{x_n}$ about a point $c$ is defined as $\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - c} \right)}^2}} $
The mean square deviation about $ - 2$ and 2 are 18 and 10 respectively, then standard deviation of this set of observations is
A.3
B.2
C.1
D.None of these

Answer 1

Hint: Use the formula for the mean square deviation to form equations for points $ - 2$and 2. Use the formed equations to find the value of \[\sum\limits_{i = 1}^n {{x_i}^2} \] and $\sum\limits_{i = 1}^n {{x_i}} $ in terms of $n$ for the set of observations. Substitute these values in the formula for standard deviation $\sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2} - {{\left( {\dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} $ to calculate the standard deviations.

Complete step-by-step answer:
Let there be $n$observations in the set. Since the mean square deviation of set of $n$observations ${x_1},{x_2}....{x_n}$ about a point $c$ is defined as $\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - c} \right)}^2}} $, and we are given that the mean square deviation about $ - 2$ is 18. Thus replacing $ - 2$ by $c$, we get
$\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \left( { - 2} \right)} \right)}^2}} = 18$
On simplifying,
$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 2} \right)}^2}} = 18n{\text{ (1)}}$
Similarly, we are also given that the mean square deviation around 2 is 10.
Thus, $\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \left( 2 \right)} \right)}^2}} = 10$
$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 2} \right)}^2}} = 10n{\text{ (2)}}$
On adding equations 1 and 2, we get
$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 2} \right)}^2}} + \sum\limits_{i = 1}^n {{{\left( {{x_i} - 2} \right)}^2}} = 10n + 18n{\text{ }}$
We can simplify the above expression as:
$
  \sum\limits_{i = 1}^n {{{\left( {{x_i} + 2} \right)}^2}} + \sum\limits_{i = 1}^n {{{\left( {{x_i} - 2} \right)}^2}} = 28n \\
  \sum\limits_{i = 1}^n {{{\left( {{x_i} + 2} \right)}^2} + {{\left( {{x_i} - 2} \right)}^2}} = 28n \\
  \sum\limits_{i = 1}^n {{x_i}^2 + 4{x_i} + 4 + {x_i}^2 - 4{x_i} + 4} = 28n \\
  \sum\limits_{i = 1}^n {2{x_i}^2 + 8} = 28n \\
  2\sum\limits_{i = 1}^n {{x_i}^2 + 4} = 28n \\
  \sum\limits_{i = 1}^n {{x_i}^2 + 4} = 14n \\
  \sum\limits_{i = 1}^n {{x_i}^2 + \sum\limits_{i = 1}^n 4 } = 14n \\
  \sum\limits_{i = 1}^n {{x_i}^2} + 4n = 14n \\
  \sum\limits_{i = 1}^n {{x_i}^2} = 10n \\
$
On subtracting equation 2 from 1, we get
$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 2} \right)}^2}} - \sum\limits_{i = 1}^n {{{\left( {{x_i} - 2} \right)}^2}} = 18n - 10n$
We can simplify the above expression as:
$
  \sum\limits_{i = 1}^n {{{\left( {{x_i} + 2} \right)}^2} - {{\left( {{x_i} - 2} \right)}^2}} = 8n \\
  \sum\limits_{i = 1}^n {\left( {{x_i} + 2 + {x_i} - 2} \right)\left( {{x_i} + 2 - {x_i} + 2} \right)} = 8n \\
  \sum\limits_{i = 1}^n {\left( {2{x_i}} \right)\left( 4 \right)} = 8n \\
  8\sum\limits_{i = 1}^n {{x_i}} = 8n \\
  \sum\limits_{i = 1}^n {{x_i}} = n \\
 $
The standard deviation of set of $n$observations ${x_1},{x_2}....{x_n}$ is defined as $\sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2} - {{\left( {\dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} $
Substituting $10n$ for \[\sum\limits_{i = 1}^n {{x_i}^2} \] and $n$ for $\sum\limits_{i = 1}^n {{x_i}} $ in the equation $\sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2} - {{\left( {\dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} $, we get
$
  \sigma = \sqrt {\dfrac{{1\left( {10n} \right)}}{n} - {{\left( {\dfrac{{1\left( n \right)}}{n}} \right)}^2}} \\
  \sigma = \sqrt {10 - {{\left( 1 \right)}^2}} \\
  \sigma = \sqrt 9 \\
  \sigma = 3 \\
 $
Hence the standard deviation of the set of $n$observations is 3.
Thus option A is the correct answer.

Note: The standard deviation of the set of $n$ observations ${x_1},{x_2}....{x_n}$ is defined as $\sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2} - {{\left( {\dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} $ and the mean square deviation of set of $n$observations ${x_1},{x_2}....{x_n}$ about a point $c$ is defined as $\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - c} \right)}^2}} $. Also the value \[\sum\limits_{i = 1}^n 4 \] equals $4n$ as \[\sum\limits_{i = 1}^n 4 = 4 + 4 + 4 + .......4\;n{\text{ times}}\].