Question

Find the mean and variance of the first 10 multiples of 3.

Answer 1

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 $\bar x = \dfrac{{\sum {{x_i}} }}{n}$ and then the mean of squares of given data values $\overline {{x^2}} = \dfrac{{\sum {{x_i}^2} }}{n}$. After that, we find the variance as ${\sigma ^2} = \overline {{x^2}} - {\left( {\bar x} \right)^2}$.

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 ${x_1},{x_2}, \ldots ,{x_n}$ then the mean of the data sample is given by,
$\bar x = \dfrac{{\sum {{x_i}} }}{n}$
Variance is the mean of squared deviations of the sample mean. It is given by the formula,
${\sigma ^2} = \overline {{x^2}} - {\left( {\bar x} \right)^2}$
Here $\overline {{x^2}}$ is the mean of squares of data values and given by,
$\overline {{x^2}} = \dfrac{{\sum {{x_i}^2} }}{n}$
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 the first 10 multiples of 3 which means the data sample is,
$\Rightarrow 3,6,9,12,15,18,21,24,27,30$
Here the number of data values is $n = 10$.
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,
$\Rightarrow \sum {{x_i}} = 3 + 6 + 9 + \ldots + 30$
We take 3 commons on the right side of the equation and have
$\Rightarrow \sum {{x_i}} = 3\left( {1 + 2 + 3 + \ldots + 10} \right)$
We use the formula for the sum of first $n$ natural numbers for $n = 10$ on the right side of the equation,
$\Rightarrow \sum {{x_i}} = 3 \times \dfrac{{10 \times 11}}{2}$
Simplify the terms,
$\Rightarrow \sum {{x_i}} = 165$
Substitute the value in the mean formula,
$\Rightarrow \bar x = \dfrac{{165}}{{10}}$
Divide numerator by the denominator,
$\Rightarrow \bar x = 16.5$...............….. (1)
Let us find the mean of squared difference of each data value from the mean
$\Rightarrow \sum {{x_i}^2} = {3^2} + {6^2} + {9^2} + \ldots + {30^2}$
We take ${3^2}$ common on the right side of the equation and have
$\Rightarrow \sum {{x_i}^2} = {3^2}\left( {{1^2} + {2^2} + {3^2} + \ldots + {{10}^2}} \right)$
We use the formula for the sum of first squared $n$ natural numbers for $n = 10$ on the right side of the equation,
$\Rightarrow \sum {{x_i}^2} = {3^2} \times \dfrac{{10 \times 11 \times 21}}{6}$
Simplify the terms,
$\Rightarrow \sum {{x_i}^2} = 3465$
Substitute the value in the mean formula,
$\Rightarrow \overline {{x^2}} = \dfrac{{3465}}{{10}}$
Divide numerator by the denominator,
$\Rightarrow \overline {{x^2}} = 346.5$...................….. (2)
Substitute the values from equation (1) and (2) in the variance formula,
$\Rightarrow {\sigma ^2} = 346.5 - {\left( {16.5} \right)^2}$
Square the term on the right side,
$\Rightarrow {\sigma ^2} = 346.5 - 272.25$
Simplify the term,
$\therefore {\sigma ^2} = 74.25$

Hence, the mean is $16.5$ and the variance is $74.25$.

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