Question

A group of $50$ items has mean $70$ and standard deviation $8$. What will be the value of mean and standard deviation if each value is multiplied by $5$?

Answer 1

Hint: We are given that a group of $50$ items has mean $70$ and standard deviation $8$ and we need to find the mean and standard deviation when each value is multiplied by $5$.
Mean is the sum of the values of all observations divided by the total number of observations and the standard deviation is the positive square root of the variance of a variance $X$. Standard deviation is denoted by $\sigma $.
We can calculate the mean and standard deviation of values if each value is multiplied by $5$, by simply multiplying each observation by $5$ and then comparing the new mean and standard deviation to the old ones.
Formula:
Mean: $\overline X = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_{50}}}}{n}$
Standard deviation: $\sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {{{({X_i} - \bar X)}^2}} } $

Complete step-by-step answer:
We have,
$n = 50$ ($n$ = no. of observations)
Mean $\left( {\overline X } \right) = 70$
Standard deviation $\left( \sigma \right) = 8$
Calculations for mean:
Let the observations be ${x_1},{x_2},{x_3},....,{x_{50}}$ , $n = 50$ and $\overline X $ be their mean. Mean $\left( {\overline X } \right) = 70$.
$\overline X = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_{50}}}}{n}$
$ \Rightarrow 70 = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_{50}}}}{{50}}$ $.....\left( i \right)$
If each value is multiplied by $5$, then the new observations be:
${x_1} \times 5,{x_2} \times 5,{x_3} \times 5,....,{x_{50}} \times 5$
$ \Leftrightarrow 5{x_1},5{x_2},5{x_3},....,5{x_{50}}$
Let the new observations be $5{x_1},5{x_2},5{x_3},....,5{x_{50}}$, $n = 50$ and $\overline X '$ be their mean or new mean.
\[ \Rightarrow \overline X ' = \dfrac{{5{x_1} + 5{x_2} + 5{x_3} + .... + 5{x_{50}}}}{{50}}\]
Take $5$ as common
$ \Rightarrow \overline X ' = \dfrac{{5\left( {{x_1} + {x_2} + {x_3} + ...... + {x_{50}}} \right)}}{{50}}$
It can also be written as:
$ \Rightarrow \overline X ' = 5\dfrac{{\left( {{x_1} + {x_2} + {x_3} + ...... + {x_{50}}} \right)}}{{50}}$
(We know that $70 = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_{50}}}}{{50}}$, from $\left( i \right)$ )
$ \Rightarrow \overline X ' = 5 \times \overline X $ or $\overline X ' = 5 \times 70$
\[ \Rightarrow \overline X ' = 350\]
Thus new mean or value of mean if each value is multiplied by $5$ is $350$.
Calculations for standard deviation:
Let the observations be ${x_1},{x_2},{x_3},....,{x_{50}}$ , $n = 50$, $\overline X $ be their mean and Standard deviation $\left( \sigma \right) = 8$
$\sigma = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {{{({X_i} - \bar X)}^2}} } $
On squaring both sides, we get
$ \Rightarrow {\sigma ^2} = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{({X_i} - \bar X)}^2}} $
Substitute value of $\sigma $ and no. of observations.
$ \Rightarrow {8^2} = \dfrac{1}{n}\sum\limits_{i = 1}^{50} {{{({X_i} - \bar X)}^2}} $
$ \Rightarrow 64 = \dfrac{1}{n}\sum\limits_{i = 1}^{50} {{{({X_i} - \bar X)}^2}} $ …..$\left( {ii} \right)$
If each value is multiplied by $5$, new observations will be: ${x_1} \times 5,{x_2} \times 5,{x_3} \times 5,....,{x_{50}} \times 5$
$ \Leftrightarrow 5{x_1},5{x_2},5{x_3},....,5{x_{50}}$
Let the new observations be $5{x_1},5{x_2},5{x_3},....,5{x_{50}}$, $n = 50$, $\overline X '$ be their mean or new mean and $\sigma '$ be the new standard deviation.
$ \Rightarrow \sigma ' = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {{{(X{'_i} - \bar X')}^2}} } $
On squaring both sides, we get
$ \Rightarrow {\left( {\sigma '} \right)^2} = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{(X{'_i} - \bar X')}^2}} $
As we know no. of observations are fixed, so put $n = 50$.
$ \Rightarrow {\left( {\sigma '} \right)^2} = \dfrac{1}{n}\sum\limits_{i = 1}^{50} {{{(X{'_i} - \bar X')}^2}} $
As we know $\overline X ' = 5 \times \overline X $
$ \Rightarrow {\left( {\sigma '} \right)^2} = \dfrac{1}{n}\sum\limits_{i = 1}^{50} {{{(X{'_i} - 5\bar X)}^2}} $
On expanding, we get
$ \Rightarrow {\left( {\sigma '} \right)^2} = \dfrac{1}{n}\left[ {{{\left( {{X_1}' - 5\overline X } \right)}^2} + {{\left( {{X_2}' - 5\overline X } \right)}^2} + ...... + {{\left( {{X_{50}}' - 5\overline X } \right)}^2}} \right]$
As we know every observation is multiplied by $5$ so we can write ${X_1}'$ as $5{X_1}$ , we will repeat this for every observation.
$ \Rightarrow {\left( {\sigma '} \right)^2} = \dfrac{1}{n}\left[ {{{\left( {5{X_1} - 5\overline X } \right)}^2} + {{\left( {5{X_2} - 5\overline X } \right)}^2} + ...... + {{\left( {5{X_{50}} - 5\overline X } \right)}^2}} \right]$
Take ${5^2}$ common from every observation.
$ \Rightarrow {\left( {\sigma '} \right)^2} = \dfrac{1}{n}\left[ {{5^2}{{\left( {{X_1} - \overline X } \right)}^2} + {5^2}{{\left( {{X_2} - \overline X } \right)}^2} + ...... + {5^2}{{\left( {{X_{50}} - \overline X } \right)}^2}} \right]$
Take ${5^2}$ as common, from the bracket
$ \Rightarrow {\left( {\sigma '} \right)^2} = \dfrac{1}{n} \times {5^2}\left[ {{{\left( {{X_1} - \overline X } \right)}^2} + {{\left( {{X_2} - \overline X } \right)}^2} + ...... + {{\left( {{X_{50}} - \overline X } \right)}^2}} \right]$
$ \Rightarrow {\left( {\sigma '} \right)^2} = 25 \times \dfrac{1}{n}\left[ {{{\left( {{X_1} - \overline X } \right)}^2} + {{\left( {{X_2} - \overline X } \right)}^2} + ...... + {{\left( {{X_{50}} - \overline X } \right)}^2}} \right]$
On contraction, we get
$ \Rightarrow {\left( {\sigma '} \right)^2} = 25 \times \dfrac{1}{n}{\left( {\sum\limits_{i = 1}^{50} {{X_i} - \overline X } } \right)^2}$
As we know ${\sigma ^2} = \dfrac{1}{n}\sum\limits_{i = 1}^{50} {{{({X_i} - \bar X)}^2}} $ from $\left( {ii} \right)$
$ \Rightarrow {\left( {\sigma '} \right)^2} = 25 \times {\left( \sigma \right)^2}$
Substitute the value of ${\sigma ^2} = 64$
$ \Rightarrow {\left( {\sigma '} \right)^2} = 25 \times 64$
On multiplying, we get
$ \Rightarrow {\left( {\sigma '} \right)^2} = 1600$
Take square roots on both the sides
$ \Rightarrow \sigma ' = \sqrt {1600} $
$ \Rightarrow \sigma ' = 40$
Thus, the mean and standard deviation if each value is multiplied by $5$ is $350$ and $40$.

Note: We wrote only the positive value of standard deviation as our answer because standard deviation is the positive square root of the variance of a variate $X$. Don’t forget to multiply observations by $5$. We should take care of the calculations so as to be sure of our final answer.