Question

The sum and sum of squares corresponding to length $X$ (in cm ) and weight $y$
(in gm) of $50$ plant products are given below:
$\sum _{i=l}^{50}{X}_i=212,\sum _{i=l}^{50}{X}_i^2=902.8,\sum _{i=l}^{50}{y}_i=261,\sum _{i=l}^{50}{y}_i^2=1457.6$
Which is more varying the length or weight ?

Answer 1

$\sum _{i=1}^{50}{X}_i=212,\sum _{i=1}^{50}{X}_i^2=902.8,$
Here, $N=50$

Mean $\overline{X_i}=\frac{{\sum }_{i=1}^{50}}{N}=\frac{212}{50}=4.24$

Variance $\left({\sigma }_1^2\right)=\frac{1}{N}\sum _{i=1}^{50}{(X_i-\overline{X})}^2$
$=\frac{1}{50}\sum _{i=1}^{50}(X_i-4.24{)}^2$
$=\frac{1}{50}\sum _{i=1}^{50}[X_i^2-8.48X_i+17.97]$
$=\frac{1}{50}[\sum _{i=1}^{50}{X}_i^2-8.48\sum _{i=1}^{50}{X}_i+17.97\times 50]$
$=\frac{1}{50}[902.8-8.48\times (212)+898.5]$
$=\frac{1}{50}[1801.3-1797.76]$
$=\frac{1}{50}\times 3.54$
$=0.07$

Also, given $\sum _{i=1}^{50}{y}_i=261,\sum _{i=1}^{50}{y}_i^2=1457.6$

Mean $\overline{y}=\frac{1}{N}\sum _{i=1}^{50}{y}_i=\frac{1}{50}\times 261=5.22$

Variance $\left({\sigma }_2^2\right)=\frac{1}{N}\sum _{i=1}^{50}{(y_i-\overline{y})}^2$
$=\frac{1}{50}\sum _{i=1}^{50}(y_i-5.22{)}^2$
$=\frac{1}{50}\sum _{i=1}^{50}[y_1^2-10.44y_i+27.24]$
$=\frac{1}{50}[\sum _{i=1}^{50}{y}_i^2-10.44y_i+27.24\times 50]$
$=\frac{1}{50}[1457.6-10.44\times (261)+1362]$
$=\frac{1}{50}[2819.6-2724.84]$
$=\frac{1}{50}\times 94.76$
$=1.89$
Since, variance of weights ($y_i$) is greater than the variance of lengths ($X_i$)

i.e ${\sigma }_2>{\sigma }_1$

So, weight vary more than the lengths .