Question

If each of the data set is increased by 10% what happens to the value of the
Mean
Mean deviation
Standard deviation

Answer 1

Let data points as $x_i$ and number of data points as n.

After increase every data$(x_i)$ point by 10%,new data points$(y_i)$ are,

$y_i=x_i+x_i*0.1=1.1*x_i$

mean of data set before increment$(\bar{x})$ ,

$\bar{x}=\frac{\sum\limits_{i=1}^{n}{x_i}}{n}\\$

mean of the data set after increment$(\bar{y})$ ,

$\bar{y}=\frac{\sum\limits_{i=1}^{n}{y_i}}{n}\\ \bar{y}=\frac{\sum\limits_{i=1}^{n}{1.1*x_i}}{n}\\ \bar{y}=1.1\frac{\sum\limits_{i=1}^{n}{x_i}}{n}\\ \bar{y}=1.1\bar{x}$

Therefore mean is also increment by 10% when data set increment by 10%.

mean deviation of the data set before increment$(M_x),$

$M_x=\frac{\sum \limits_{i=1}^{n}{\text{\textbar} x_i-\bar{x}\text{\textbar}}}{n}$

mean deviation of the data set before increment$(M_y),$

$M_y=\frac{\sum \limits_{i=1}^{n}{\text{\textbar} y_i-\bar{y}\text{\textbar}}}{n}\\ M_y=\frac{\sum \limits_{i=1}^{n}{\text{\textbar}1.1* x_i-1.1*\bar{x}\text{\textbar}}}{n}\\ M_y=\frac{1.1*\sum \limits_{i=1}^{n}{\text{\textbar} x_i-\bar{x}\text{\textbar}}}{n}\\ M_y=1.1*M_x$

Therefore mean deviation is also increment by 10% when data set increment by 10%.

variance of the data set before increment$(\sigma^2_x)$

$\sigma^2_x=\frac{\sum\limits_{i=1}^{n}{(x_i-\bar{x})^2}}{n}=\frac{\sum\limits_{i=1}^{n}{x_i^2}}{n}-(\bar{x})^2\\$

variance of the data set after increment$(\sigma^2_y)$

$\sigma^2_y=\frac{\sum\limits_{i=1}^{n}{y_i^2}}{n}-(\bar{y})^2\\ \sigma^2_y=\frac{\sum\limits_{i=1}^{n}{(1.1*x_i)^2}}{n}-(1.1*\bar{x})^2\\ \sigma^2_y=(1.1)^2*\frac{\sum\limits_{i=1}^{n}{(x_i)^2}}{n}-(1.1)^2*(\bar{x})^2\\ \sigma^2_y=(1.1)^2*\sigma^2_x\\ \sigma^2_y=1.21*\sigma^2_x$

standard deviation,

$\sigma_y=1.1*\sigma_x$

Therefore standard deviation is also increment by 10% when data set increment by 10%.