Question

Find the mean and variance of the first 10 multiples of 3. \[\]

Answer 1

Hint: We find the mean of the given data sample by first finding the sum of data values then dividing by the number of data values which in symbols is $ \overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}$ and then mean of squares of given data values $ \overline{{{x}^{2}}}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}}$. We find the variance as $ {{\sigma }^{2}}=\overline{{{x}^{2}}}-{{\left( \overline{x} \right)}^{2}}$.\[\]

Complete step-by-step solution:
We know that the mean is the expectation or average of the given data value. If there are $n$ data values say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ then mean of data sample is given by
\[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\]
Variance is the mean of squared deviations of sample mean. It is given by the formula
\[{{\sigma }^{2}}=\overline{{{x}^{2}}}-{{\left( \overline{x} \right)}^{2}}\]
Here $\overline{{{x}^{2}}}$ the mean of squares of data values and given by
\[\overline{{{x}^{2}}}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}}\]
We know that the sum $S$ of first $n$ natural numbers is given by
\[S=\dfrac{n\left( n+1 \right)}{2}\]
We know that the sum ${{S}_{2}}$ of first squared $n$ natural numbers is given by,
\[{{S}_{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
From the question we are given the data sample of first 10 multiples of 3 which means the data sample is
\[\begin{align}
  & 3\times 1,3\times 2,3\times 3,3\times 4,3\times 5,3\times 6,3\times 7,3\times 8,3\times 9,3\times 10 \\
 & =3,6,9,12,15,18,21,24,26,30 \\
\end{align}\]
Here number of data values is $n=10.$Let us denote the data values as
\[{{x}_{1}}=3,{{x}_{2}}=6,{{x}_{3}}=9,...,{{x}_{10}}=30\]
Let us first find the mean of the data sample. So we need to find the sum of data values. We have to first the sum of data values which is
\[\begin{align}
  & \sum\limits_{i=1}^{n}{{{x}_{i}}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}}...+{{x}_{10}} \\
 & \Rightarrow \sum\limits_{i=1}^{n}{{{x}_{i}}}=3+6+9...+30 \\
\end{align}\]
We take 3 common in the right hand side of the equation and have
\[\Rightarrow \sum\limits_{i=1}^{n}{{{x}_{i}}}=3\left( 1+2+3...10 \right)\]
We use the formula for the sum of first $n$ natural numbers for $n=10$in the right hand side of the equation and have,
\[\Rightarrow \sum\limits_{i=1}^{n}{{{x}_{i}}}=3\left( \dfrac{10\times 11}{2} \right)=165\]
So the sample mean is
\[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}=\dfrac{165}{10}=16.5\]
Let us find the mean of squared difference of each data value from the mean
\[\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}={{3}^{2}}+{{6}^{2}}+...+{{30}^{2}}}\]
Let us take ${{3}^{2}}$ common in the right hand side of the equation and have
\[\Rightarrow \sum\limits_{i=1}^{n}{{{x}_{i}}^{2}={{3}^{2}}\left( {{1}^{2}}+{{2}^{2}}+{{...10}^{2}} \right)}\]
We use the formula for the sum of first squared $n$ natural numbers for $n=10$in the right hand side of the equation and have,
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{10}{{{x}_{i}}^{2}=9\left( \dfrac{10\times \left( 10+1 \right)\left( 2\times 10+1 \right)}{6} \right)} \\
 & \Rightarrow \sum\limits_{i=1}^{10}{{{x}_{i}}^{2}=9\left( \dfrac{10\times 11\times 21}{6} \right)} \\
 & \Rightarrow \sum\limits_{i=1}^{10}{{{x}_{i}}^{2}=3465} \\
\end{align}\]
The mean of squared differences is
\[\overline{{{x}^{2}}}=\dfrac{1}{n}\sum\limits_{i=1}^{10}{{{x}_{i}}^{2}}=\dfrac{1}{10}\times 3465=346.5\]
So the sample variance is
\[{{\sigma }^{2}}=\overline{{{x}^{2}}}-{{\left( \overline{x} \right)}^{2}}=346.5-{{\left( 16.5 \right)}^{2}}=346.5-272.5=74.25\]

Note: We note that the variance is always a positive quantity while mean may not be. The squareroot of variance is called the standard deviation $\sigma $ and the ratio of standard deviation to mean $\overline{x}$ is called coefficient variation, a useful quantity in risk analysis. We can alternatively find variance with the formula $\dfrac{1}{n}{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}$.