Question

The variance of observations 112, 116, 120, 125,132 is \[\]
A. 58.8\[\]
B.48.8\[\]
C. 61.8\[\]
D. None of these \[\]

Answer 1

Hint: We first find the mean of the given data sample 112, 116, 120, 125,132 by using the formula $\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}$ where $n$ is the number of data values and ${{x}_{i}}$’s are the data values. We find the squared deviation from the mean as ${{\left( {{x}_{i}}-\overline{x} \right)}^{2}}$ and take the mean of squared deviations to find the variance. \[\]

Complete step by step answer:
We know that mean is the expectation or average of the given data value. If they are $n$ data values say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ then mean of data sample is denoted by $\overline{x}$ and given by
\[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\]
We also know that variance is the mean of squared deviation from the mean. The deviation of any data value ${{x}_{i}},i=1,2,...,n$ from the mean $\overline{x}$ is given by $\left( {{x}_{i}}-\overline{x} \right)$. The square of the deviation is ${{\left( {{x}_{i}}-\overline{x} \right)}^{2}}$. The mean of all such squared deviations is the variance which is denoted by ${{\sigma }^{2}}$ and is given by
\[{{\sigma }^{2}}=\dfrac{\sum\limits_{i=n}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}\]
We observe the given data in the question 112, 116, 120, 125,132. We count the data values and find the number of data values as $n=5$. We can denote the data values as
\[{{x}_{1}}=112,{{x}_{2}}=116,{{x}_{3}}=120,{{x}_{4}}=125,{{x}_{5}}=132\]
The mean of the data sample is

\[\overline{x}=\dfrac{1}{5}\sum\limits_{i=1}^{5}{{{x}_{i}}}=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}}{5}=\dfrac{112+116+120+125+132}{5}=\dfrac{605}{5}=121\]
Let us find the squares of deviations from the mean. We have,
\[\begin{align}
  & {{\left( {{x}_{1}}-\overline{x} \right)}^{2}}={{\left( 112-121 \right)}^{2}}={{\left( -9 \right)}^{2}}=81 \\
 & {{\left( {{x}_{2}}-\overline{x} \right)}^{2}}={{\left( 116-121 \right)}^{2}}={{\left( -5 \right)}^{2}}=25 \\
 & {{\left( {{x}_{3}}-\overline{x} \right)}^{2}}={{\left( 120-121 \right)}^{2}}={{\left( -1 \right)}^{2}}=1 \\
 & {{\left( {{x}_{4}}-\overline{x} \right)}^{2}}={{\left( 125-121 \right)}^{2}}={{\left( 4 \right)}^{2}}=16 \\
 & {{\left( {{x}_{5}}-\overline{x} \right)}^{2}}={{\left( 132-121 \right)}^{2}}={{\left( 11 \right)}^{2}}=121 \\
\end{align}\]
The mean of the squares of deviations from the mean of data sample is
\[{{\sigma }^{2}}=\dfrac{\sum\limits_{i=n}^{5}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{5}=\dfrac{81+25+1+16+121}{5}=\dfrac{244}{5}=48.8\]


So, the correct answer is “Option B”.

Note: We note that the variance is always a positive quantity while mean may not be. The square root of variance is called standard deviation $\sigma $ and the ratio of standard deviation to mean $\overline{x}$ is called coefficient variation, a useful quantity in the risk analysis. The variance is always measured in squared units of the data value and signifies the spread of the data value from the mean.