Question

The sum and the sum of squares of length x (in cm) and weight y (in g) of 50 plant products are given below:

${\sum }_{i=1}^{50}{x}_i=212,{\sum }_{i=1}^{50}{x}_i^2=902.8,{\sum }_{i=1}^{50}{y}_i=261$ and ${\sum }_{i=1}^{50}{y}_i^2=1457.6$

Which is more variable, the length or weight?

Answer 1

$var\left(x\right)=\frac{1}{50}.{\sum }_{i=1}^{50}{x}_i^2-{(\frac{1}{50}.{\sum }_{i=1}^{50}{x}_i)}^2=\left(\frac{902.8}{50}\right)-{\left(\frac{212}{50}\right)}^2=\frac{451.4}{25}-{\left(\frac{106}{25}\right)}^2$

$=(\frac{451.4}{25}-\frac{11236}{625})=\left(\frac{11285-11236}{625}\right)$

$=\frac{49}{625}=0.078$

$var\left(y\right)=\frac{1}{50}.{\sum }_{i=1}^{50}{y}_i^2-{(\frac{1}{50}.{\sum }_{i=1}^{50}{y}_i)}^2=\{\frac{1457.6}{50}-{\left(\frac{261}{50}\right)}^2\}$

$=\{29.152-(5.22{)}^2\}=29.152-27.2484$

= 1.9036

$\therefore$ var(x) < var (y)