Question

A data consists of \[n\] observations: \[{x_1},{x_{2,...,}}{x_{n.}}\]
If, \[\sum\limits_{i = 1}^n {{{({x_i} + 1)}^2}} = 9n\] and \[\sum\limits_{i = 1}^n {{{({x_i} - 1)}^2}} = 5n\], then find the standard deviation.

Answer 1

Hint:
We will compare the equations given in the equation and reduce the left hand side of the equation as variance, since we know that the square root of variance is standard deviation we will find the value.

Formula used:
The Standard Deviation is a measure of how spread out numbers are.
The standard deviation of \[n\] observations:\[{x_1},{x_{2,...,}}{x_{n.}}\]is:
\[\sqrt {\sum\limits_{i = 1}^n {\dfrac{{{{({x_i} - \mu )}^2}}}{n}} } \], where, \[\mu \] is the known as population mean.
In other words, the standard variance is the square root of variance.
The variance of \[n\] observations:\[{x_1},{x_{2,...,}}{x_{n.}}\]is:
\[\sum\limits_{i = 1}^n {\dfrac{{{x_i}^2 - {x_i}}}{n}} \]

Complete step-by-step answer:
It is given that there are \[n\] observations:\[{x_1},{x_{2,...,}}{x_{n.}}\]
Also, it is given that,
\[\sum\limits_{i = 1}^n {{{({x_i} + 1)}^2}} = 9n\]…… (1)
\[\sum\limits_{i = 1}^n {{{({x_i} - 1)}^2}} = 5n\]….. (2)
Let us now add equation (1) and (2), so that we get the following equation,
\[\sum\limits_{i = 1}^n {{{({x_i} + 1)}^2}} + \sum\limits_{i = 1}^n {{{({x_i} - 1)}^2}} = 9n + 5n\]
Now let us expand the summations in the left hand side of the equation and regroup it so that we get the equation below,
\[({x_1}^2 + 2{x_1} + 1 + {x_1}^2 - 2{x_1} + 1) + ({x_2}^2 + 2{x_2} + 1 + {x_2}^2 - 2{x_2} + 1) + ... + ({x_n}^2 + 2{x_n} + 1 + {x_n}^2 - 2{x_n} + 1) = 14n\]
Let us simplify the above equation and write it in the summation form we get,
\[2(\sum\limits_{i = 1}^n {{x_i}^2 + n) = } 14n\]
Let us divide the above equation by 2 on both sides we get,
\[(\sum\limits_{i = 1}^n {{x_i}^2 + n) = } 7n\]
Also let us divide it both sides by \[n\]we get,
\[\sum\limits_{i = 1}^n {\dfrac{{{x_i}^2}}{n} + 1 = } 7\]
By further simplification we get,
\[\sum\limits_{i = 1}^n {\dfrac{{{x_i}^2}}{n} = } 6\]…(3)
Now, we will subtract the two equations and compare it with equation (3)
Subtracting equation (1) and (2) we get,
\[\sum\limits_{i = 1}^n {{{({x_i} + 1)}^2}} - \sum\limits_{i = 1}^n {{{({x_i} - 1)}^2}} = 9n - 5n\]
Here we expand the summations in the left hand side and regroup it to get the following equation,
\[({x_1}^2 + 2{x_1} + 1 - {x_1}^2 + 2{x_1} - 1) + ({x_2}^2 + 2{x_2} + 1 - {x_2}^2 + 2{x_2} - 1) + ... + ({x_n}^2 + 2{x_n} + 1 - {x_n}^2 + 2{x_n} - 1) = 4n\]
On solving the equation in the left hand side and representing it in the summation we get,
\[4\sum\limits_{i = 1}^n {{x_i} = } 4n\]
By dividing the above equation by \[4n\]on both sides we get,
\[\sum\limits_{i = 1}^n {\dfrac{{{x_i}}}{n} = } 1\]…(4)
So, the variance is found by subtracting (3) and (4), we get
\[\sum\limits_{i = 1}^n {\dfrac{{{x_i}^2 - {x_i}}}{n} = } 6 - 1 = 5\]
Since the standard deviation is the square root of variance,
The standard deviation is \[\sqrt {\sum\limits_{i = 1}^n {\dfrac{{{x_i}^2 - {x_i}}}{n} = } } \sqrt 5 \]
Hence the standard deviation is \[\sqrt 5 \]

Note:
The standard deviation is also defined as the square root of the variance. Since the variance is \[\sum\limits_{i = 1}^n {\dfrac{{{x_i}^2 - {x_i}}}{n}} \] the standard deviation is given by the formula \[\sqrt {\sum\limits_{i = 1}^n {\dfrac{{{x_i}^2 - {x_i}}}{n}} } \]