Question

The variance of the series \[3,5,8,6,12\] and \[11\] is

Answer 1

Hint: In the given question, we have to find out the variance of given series. Variance is basically a statistical measurement of the spread between the numbers in a given data set. Variance measures that your number is far from the mean of the observations where mean is the division of sum of observations to the number of observations.

Complete answer:
In the given question, we have to find out the variance of the series \[3,5,8,6,12\] and \[11\].
For this, firstly we will find mean because it will be used in the formula of variance.
\[{\text{Mean}} = \dfrac{{{\text{Sum of observations}}}}{{{\text{Number of observations}}}}\]
Where sum of observations means \[3 + 5 + 8 + 6 + 12 + 11 = 45\] and the number of observations are \[6\].
Therefore, Mean\[ = \dfrac{{45}}{6}\]
\[ = 7.5\]
We denote the mean by \[\overline x \], which is \[7.5\].
So, \[\overline x = 7.5\]
Also, there are \[6\] observations which are denoted as \[n1,n2,n3,n4,n5\] and \[n6\] for \[6\] observations in the series and in general, we denote all these by \[{n_i}\]. Formula of variance is given by
Variance\[ = \dfrac{{\sum {{{\left( {{x_i} - \overline x } \right)}^2}} }}{{n - 1}}\]
Where \[{x_i}\] is the number of observations one at a time, \[\overline x \] is mean and \[n\] is the number of observations which is \[6\].
So, we firstly calculate \[{x_i} - \overline x \] and then \[{\left( {{x_i} - \overline x } \right)^2}\] as

\[\left( {{x_i} - \overline x } \right)\]	\[{\left( {{x_i} - \overline x } \right)^2}\]
\[3 - 7.5 = - 4.5\]\[5 - 7.5 = - 2.5\]\[8 - 7.5 = 0.5\]\[6 - 7.5 = - 1.5\]\[12 - 7.5 = 4.5\]\[11 - 7.5 = 3.5\]	\[20.25\]\[6.25\]\[0.25\]\[2.25\]\[20.25\]\[12.25\]
	\[\sum {{{\left( {{x_i} - \overline x } \right)}^2} = 61.5} \]

Hence, we have calculated \[\sum {{{\left( {{x_i} - \overline x } \right)}^2}} \] where \[\sum {{{\left( {{x_i} - \overline x } \right)}^2}} \] is the sum of all the observations after calculating \[{\left( {{x_i} - \overline x } \right)^2}\] and \[{\left( {{x_i} - \overline x } \right)^2}\] is obtained by squaring of \[\left( {{x_i} - \overline x } \right)\] and \[\left( {{x_i} - \overline x } \right)\] is obtained by subtracting \[{x_i}\] from \[\overline x \] one by one.
Therefore, on substituting the value of \[\sum {{{\left( {{x_i} - \overline x } \right)}^2} = 7.5} \] in formula of variance, we get
Variance\[ = \dfrac{{\sum {{{\left( {{x_i} - \overline x } \right)}^2}} }}{{n - 1}}\]
\[ \Rightarrow \dfrac{{61.5}}{{6 - 1}} = \dfrac{{61.5}}{5}\]
And hence we got \[12.3\]

Therefore, variance of the given series is \[12.3\].

Note: If we take square root of variance, we get another statistical measurement which is standard deviation. We denote standard deviation as \[\sigma \]. It means variance is square of standard deviation where standard deviation is the measurement of amount of variance and indicates that the values are spread out over a wider range.