Question

The variance of the data 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 is :
A. 15
B. 20
C. 30
D. 33

Answer 1

Hint: First find the sum of the given data by applying the formula of sum of n terms of an A.P given as : ${{S}_{n}}=\dfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right]$. Here, ${{S}_{n}}$ is the sum of n terms, ${{T}_{1}}$ is the first term and ${{T}_{n}}$ is the ${{n}^{th}}$ term or the last term. Now, divide ${{S}_{n}}$ by n to find the mean $\left( \overline{x} \right)$ of the data. Now, apply the formula for variance given by : ${{\sigma }^{2}}=\dfrac{1}{n}\left[ \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}} \right]$ to get the answer, where ${{\sigma }^{2}}$ is the variance.

Complete step-by-step solution:
Here, we have been provided with the following set of data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 and have been asked to find the variance.
To find the variance, we need to calculate the value of the mean of the data. Now, we can clearly see that the given numbers in the data set are in A.P with the first term as 6 and the last term as 24. On counting the amount of numbers, we get that they are 10 in number. So, we have,
${{T}_{1}}=6,{{T}_{n}}=24,n=10$
Therefore, applying the formula for sum of n terms of an A.P, we get, ${{S}_{n}}=\dfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right]$. Here, ${{T}_{1}}$ is the first term and ${{T}_{n}}$ is the last term and n is the number of terms. Therefore, substituting the values of ${{T}_{1}}$, ${{T}_{n}}$ and n, we get,
$\begin{align}
  & {{S}_{n}}=\dfrac{10}{2}\left[ 6+24 \right] \\
 & \Rightarrow {{S}_{n}}=5\times 30 \\
 & \Rightarrow {{S}_{n}}=150 \\
\end{align}$
Now, we know that mean is given by the formula : $\overline{x}=\dfrac{\text{sum of all the terms}}{\text{Number of terms}}$, here $\overline{x}$ is the notation of mean.
$\Rightarrow \overline{x}=\dfrac{{{S}_{n}}}{n}$
Substituting the values of ${{S}_{n}}$ and n, we get,
$\begin{align}
  & \Rightarrow \overline{x}=\dfrac{150}{10} \\
 & \Rightarrow \overline{x}=15 \\
\end{align}$
Applying the formula for variance given by : ${{\sigma }^{2}}=\dfrac{1}{n}\left[ \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}} \right]$, we get,
${{\sigma }^{2}}=\dfrac{1}{10}\left[ \sum\limits_{i=1}^{10}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}} \right]$
Here, ${{\sigma }^{2}}$ is the notation of variance. Now, substituting the value of $\overline{x}=15$, we get,
$\begin{align}
  & {{\sigma }^{2}}=\dfrac{1}{10}\times \sum\limits_{i=1}^{10}{{{\left( {{x}_{i}}-15 \right)}^{2}}} \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{10}\times \left[ {{\left( {{x}_{1}}-15 \right)}^{2}}+{{\left( {{x}_{2}}-15 \right)}^{2}}+\ldots \ldots +{{\left( {{x}_{10}}-15 \right)}^{2}} \right] \\
\end{align}$
Substituting the values of ${{x}_{1}},{{x}_{2}},\ldots \ldots ,{{x}_{10}}$ from the given question, we get,
$\begin{align}
  & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{10}\times \left[ {{\left( 6-15 \right)}^{2}}+{{\left( 8-15 \right)}^{2}}+\ldots \ldots +{{\left( 24-15 \right)}^{2}} \right] \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{10}\times \left[ {{\left( -9 \right)}^{2}}+{{\left( -7 \right)}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( -3 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+{{7}^{2}}+{{9}^{2}} \right] \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{10}\times 2\times \left[ {{1}^{2}}+{{3}^{2}}+{{5}^{2}}+{{7}^{2}}+{{9}^{2}} \right] \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{5}\times \left[ 1+9+25+49+81 \right] \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{1}{5}\times \left[ 165 \right] \\
 & \Rightarrow {{\sigma }^{2}}=33 \\
\end{align}$
Hence, option D is the correct answer.

Note: One may note that variance is nothing but the square of standard deviation. We have found the total number of terms (n) by counting because there are not many terms. If there were many terms, then we would have used the formula: ${{T}_{n}}=a+\left( n-1 \right)d$ to determine the value of n. Here, a is the first term of a given A.P and d is a common difference. We have applied the formula for the sum of n terms of A.P because it makes our calculations easy. It will be time taking to add all the numbers one-by-one.