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: $\bar{X}={\displaystyle \frac{x_1+x_2+x_3+......+x_{50}}{n}}$
Standard deviation: $\sigma =\sqrt{{\displaystyle \frac{1}{n}}\sum _{i=1}^n{(X_i-\overline{X})}^2}$

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