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

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?

 

Solution:

To find the more variable, we again need to compare the coefficients of variation (CV).

Here the number of products are n = 50 for length and weight both.

For length

Mean $=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{212}{50}=4.24$

Variance $=\frac{1}{n^{2}}\left[n \sum x_{i}^{2}-\left(\sum x_{i}\right)^{2}\right]$

$=\frac{1}{50^{2}}\left[(50 \times 902.8)-(212)^{2}\right]$

$=\frac{1}{2500}[45140-44944]$

$=\frac{196}{2500}=0.0784$

So, standard deviation, SD $=\sqrt{\text { Variance }}=\sqrt{0.0784}=0.28$

Therefore, the coefficient of variation of length

$C V_{L}=\frac{0.28}{4.24} \times 100=6.603$

For weight,

Mean $=\frac{\sum y_{i}}{n}=\frac{261}{50}=5.22$

Variance $=\frac{1}{\mathrm{n}^{2}}\left[\mathrm{n} \sum \mathrm{y}_{\mathrm{i}}^{2}-\left(\sum \mathrm{y}_{\mathrm{i}}\right)^{2}\right]$

$=\frac{1}{50^{2}}\left[(50 \times 1457.6)-(261)^{2}\right]$

$=\frac{1}{2500}[72880-68121]$

$=\frac{4759}{2500}=1.9036$

So, standard deviation, $S D=\sqrt{\text { Variance }}=\sqrt{1.9036}=1.37$

Therefore, the coefficient of variation of length,

$\mathrm{CV}_{\mathrm{w}}=\frac{1.37}{5.22} \times 100=26.245$

Now, $C V_{w}>C V_{L}$

Therefore, the weight is more variable than height.

 

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