Question

The sum and sum of square corresponding to length `x` (in cm) and weight `y` (in gm) 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`
`sum_(i=1)^(50)y_(i)^(2)=1457.6`
Which is more varying , the length or weight?

Answer 1

For length,
`sumx_(i)=212, sumx_(i)^(2)=902.8,N=50`
`:.` Mean `barx=(sumx_(i))/N=212/50=4.24cm`
Variance `=(sumx_(i)^(2))/N-((sumx_(i))/N)^(2)=902.8/50-(212/50)^(2)`
`=(45140-44944)/(50^(2))=196/2500`
`implies` Standard deviation `sigma=14/50=0.28cm`
Coefficient of variation of length `=(sigma)/x xx 100`
`=0.28/4.24xx 100=6.6`
For weight,
`sumy_(i)=261,sumy_(i)^(2)=1457.6, N=50`
`:.` Mean `=(sumy_(i))/N=261/N=5.22`
Variance `sigma^(2)=(sumy_(i)^(2))/N-((sumy_(i))/N)^(2)=1457.6/50-(261/60)^(2)`
`=(72880-68121)/((50)^(2))=4759/25500`
`implies` Standard deviation `sigma=sqrt(4759/2500)=68.98/50=1.38`
Now coefficient of variation of weight
`=(sigma)/x xx100=1.38/5.22xx 100=26.43`
Therefore, the weight of products are more varying.