Question

How do $ 2 $ calculate the variance of $ \left\{ 3,6,7,8,9 \right\} $

Answer 1

Hint: We are given a data set as $ \left\{ 3,6,7,8,9 \right\} $ . We have to find the variance. We will first define the meaning of variance then we will use $ {{S}^{2}}=\dfrac{\sum {{({{x}_{i}}-\overset{-}{\mathop{x}}\,)}^{2}}}{n-1} $ to find the value of variance to evaluate this we will first find the mean of the given data as $ \overset{-}{\mathop{x}}\, $ denoted above is mean, after this we will find the difference of each term from the mean. We calculate $ ({{x}_{i}}-\overset{-}{\mathop{x}}\,) $ then we use them in our above define formula of variance to solve further and get response solution.

Complete step by step answer:
We are given the set which contain certain term as $ \left\{ 3,6,7,8,9 \right\} $ $ $
We are asked to find variance by leaving about the variance Now variance is a term that measures how far each number in the given set is from the mean and thus from others.
We know variance is given
 $ {{S}^{2}}=\dfrac{\sum {{({{x}_{i}}-\overset{-}{\mathop{x}}\,)}^{2}}}{n-1} $
Where
 $ {{x}_{i}} $ - denotes observation
 $ \overset{-}{\mathop{x}}\, $ -Mean
 $ n $ -Sample Size
 $ {{s}^{2}} $ -denotes the variance
So, to find the variance we will follow the step as follow
Step: 1
We will find the mean of the given data
Step: 2
We find difference of mean term from mean that is we find $ \left( {{x}_{i}}-\overset{-}{\mathop{x}}\, \right) $
Step: 3
We will put the value obtained above and add them and obtain $ \sum {{\left( {{x}_{i}}-\overset{-}{\mathop{x}}\, \right)}^{2}} $
Step: 4
Put the value of $ \sum {{\left( {{x}_{i}}-\overset{-}{\mathop{x}}\, \right)}^{2}} $ and $ n-1 $ in our formula and calculate variance.
Now as we have our set $ \left\{ 3,6,7,8,9 \right\} $ clearly the set have $ 5 $ terms so sample size is $ 5 $
Hence $ n=5 $
 $ \dfrac{\text{Sum}\,\text{total}\,\text{of}\,\text{all}\,\text{observation}}{\text{Total}\,\text{number}\,\text{of}\,\text{observstion}\,} $
so,
 $ \text{Mean= }\dfrac{3+6+7+8+9}{5} $
 $ =6.6 $
So,
 $ \overset{-}{\mathop{x}}\,=6.6 $
Now we will find the difference of value from mean
 $ \begin{align}
  & \left( {{x}_{1}}-\overset{-}{\mathop{x}}\, \right)=\left( 3-6.6 \right)=-3.6 \\
 & \left( {{x}_{2}}-\overset{-}{\mathop{x}}\, \right)=\left( 6-6.6 \right)=-0.6 \\
 & \left( {{x}_{3}}-\overset{-}{\mathop{x}}\, \right)=\left( 7-6.6 \right)=0.4 \\
 & \left( {{x}_{4}}-\overset{-}{\mathop{x}}\, \right)=\left( 8-6.6 \right)=1.4 \\
 & \left( {{x}_{5}}-\overset{-}{\mathop{x}}\, \right)=\left( 9-6.6 \right)=2.4 \\
\end{align} $
Now we calculate the sum of square of there term that is we find
 $ \sum {{\left( {{x}_{i}}-\overset{-}{\mathop{x}}\, \right)}^{2}} $
So, putting value of above we get.
\[\begin{align}
  & \sum\limits_{i=1}{{{\left( {{x}_{1}}-\overset{-}{\mathop{x}}\, \right)}^{2}}+}{{\left( {{x}_{2}}-\overset{-}{\mathop{x}}\, \right)}^{2}}{{\left( {{x}_{3}}-\overset{-}{\mathop{x}}\, \right)}^{2}}+{{\left( {{x}_{4}}-\overset{-}{\mathop{x}}\, \right)}^{2}}+{{\left( {{x}_{5}}-\overset{-}{\mathop{x}}\, \right)}^{2}} \\
 & ={{\left( -3.6 \right)}^{2}}+{{\left( -0.6 \right)}^{2}}+{{\left( 0.4 \right)}^{2}}+{{\left( 1.4 \right)}^{2}}+{{\left( 2.4 \right)}^{2}} \\
\end{align}\]
Simplifying we get
\[12.96+0.36+0.16+1.96+5.76=21.2~\]
So, we finally get
Now we apply
\[{{\sum\limits_{i=1}{\left( {{x}_{i}}-\overset{-}{\mathop{x}}\, \right)}}^{2}}=21.2\,\,\,\,\text{and}\,\,\,n=5\]
\[\begin{align}
  & {{S}^{2}}=\dfrac{\sum \left( {{x}_{i}}-\overset{-}{\mathop{x}}\, \right)}{n-1} \\
 & {{S}^{2}}=\dfrac{\left( 21.2 \right)}{5-1}=\dfrac{21.2}{4} \\
\end{align}\]
So solving we get
Using these values in the below eqn, we get
\[{{S}^{2}}=5.3\]
Hence variance is \[{{S}^{2}}=5.3\].

Note:
From this point, if we take the square root of the variance, we will get the standard deviation. it tells us about how values are changing from the mean. Always remember that when we subtract terms, we need to be careful about the sign of the terms. There is one thing that value obtain from \[{{x}_{i}}-\overset{-}{\mathop{x}}\,\] their sum must always be zero if the sum is not zero that value is incorrect.