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Question: The variance of observations 112, 116, 120, 125,132 is \[\] A. 58.8\[\] B.48.8\[\] C. 61.8\[\]...

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

Explanation

Solution

We first find the mean of the given data sample 112, 116, 120, 125,132 by using the formula x‾=1n∑i=1nxi\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}} where nn is the number of data values and xi{{x}_{i}}’s are the data values. We find the squared deviation from the mean as (xi−x‾)2{{\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 nn data values say x1,x2,...,xn{{x}_{1}},{{x}_{2}},...,{{x}_{n}} then mean of data sample is denoted by x‾\overline{x} and given by
x‾=1n∑i=1nxi\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 xi,i=1,2,...,n{{x}_{i}},i=1,2,...,n from the mean x‾\overline{x} is given by (xi−x‾)\left( {{x}_{i}}-\overline{x} \right). The square of the deviation is (xi−x‾)2{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}. The mean of all such squared deviations is the variance which is denoted by σ2{{\sigma }^{2}} and is given by
σ2=∑i=nn(xi−x‾)2n{{\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=5n=5. We can denote the data values as
x1=112,x2=116,x3=120,x4=125,x5=132{{x}_{1}}=112,{{x}_{2}}=116,{{x}_{3}}=120,{{x}_{4}}=125,{{x}_{5}}=132
The mean of the data sample is

x‾=15∑i=15xi=x1+x2+x3+x4+x55=112+116+120+125+1325=6055=121\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,

& {{\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{aligned}$$ The mean of the squares of deviations from the mean of data sample is $${{\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.