Question

The average of 2, 4, 6, 8 and 10 is \[\]
A.5\[\]
B.6\[\]
C.7\[\]
D.8\[\]

Answer 1

Hint: We first find the sum of the given data values 2, 4, 6, 8 and 10 as $S=\sum\limits_{i=1}^{n}{{{x}_{i}}}$ and count the data values to find the number of data values by counting as $n=5.$ We divide the sum $S$by the number of data values $n$ to find the mean. \[\]

Complete step-by-step solution:
We know that mean or average is the expected value of the data sample set. It is the value we expect without going detailed into data and it is a measure of central tendency. It is denoted by $\overline{x}$. If there are $n$ number of data values with equal weights in the samples say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ then the mean is calculated by first finding the sum of data values and then by dividing the sum by $n.$So the sample is given by
\[\overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}=\dfrac{S}{n}\]
If the weights of data values are not equal and are assigned weights say ${{w}_{1}},{{w}_{2}},...,{{w}_{n}}$ then the sample mean is given by
\[\overline{x}=\dfrac{{{w}_{1}}{{x}_{1}}+{{w}_{2}}{{x}_{2}}+...+{{w}_{n}}{{x}_{n}}}{n}=\dfrac{\sum\limits_{i=1}^{n}{{{w}_{i}}{{x}_{i}}}}{n}\]
Sample mean is different from population mean for which number of data values is very large. It is denoted by $\mu $.If the data values occur ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ occur at probabilities ${{p}_{1}},{{p}_{2}},...,{{p}_{n}}$ then population man is given by
\[\mu =\sum{{{x}_{i}}{{p}_{i}}}\]
Let us observe the given data sample in the question 2, 4, 6, 8 and10. If we count the number of data values in the question and find to be 5. So $n=5$. We can denote the data values as
\[{{x}_{1}}=2,{{x}_{2}}=4,{{x}_{3}}=6,{{x}_{4}}=8,{{x}_{5}}=10\]
Let us find the sum of the data values. We have
\[\begin{align}
  & S=\sum\limits_{i=1}^{n}{{{x}_{i}}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}} \\
 & \Rightarrow S=2+4+6+8+10=30 \\
\end{align}\]
Let us find the average of the given data by dividing the sum of data values by the number of data values. We have the average as
\[\overline{x}=\dfrac{S}{n}=\dfrac{30}{5}=6\]
So the correct option is C.

Note: Here the mean is the arithmetic means AM. Other types of mean are geometric mean and harmonic mean. Geometric mean GM is given by ${{n}^{\text{th}}}$root of the product of data values which means ${{\left( \prod\limits_{i=1}^{n}{{{x}_{i}}} \right)}^{\dfrac{1}{n}}}$ and harmonic mean is the mean of reciprocals of data values and is given by $n{{\left( \sum\limits_{i=1}^{n}{\dfrac{1}{{{x}_{i}}}} \right)}^{-1}}$. The relation among mean is $AM\ge GM\ge HM$.