The sum and sum of squares corresponding to length X (in cm ) and weight y
(in gm) of 50 plant products are given below:
50∑i=lXi=212,50∑i=lX2i=902.8,50∑i=lyi=261,50∑i=ly2i=1457.6
Which is more varying the length or weight ?
50∑i=1Xi=212,50∑i=1X2i=902.8,
Here, N=50
Mean ¯Xi=∑50i=1N=21250=4.24
Variance (σ21)=1N50∑i=1(Xi−¯X)2
=15050∑i=1(Xi−4.24)2
=15050∑i=1[X2i−8.48Xi+17.97]
=150[50∑i=1X2i−8.4850∑i=1Xi+17.97×50]
=150[902.8−8.48×(212)+898.5]
=150[1801.3−1797.76]
=150×3.54
=0.07
Also, given 50∑i=1yi=261,50∑i=1y2i=1457.6
Mean ¯y=1N50∑i=1yi=150×261=5.22
Variance (σ22)=1N50∑i=1(yi−¯y)2
=15050∑i=1(yi−5.22)2
=15050∑i=1[y21−10.44yi+27.24]
=150[50∑i=1y2i−10.44yi+27.24×50]
=150[1457.6−10.44×(261)+1362]
=150[2819.6−2724.84]
=150×94.76
=1.89
Since, variance of weights (yi) is greater than the variance of lengths (Xi)
i.e σ2>σ1
So, weight vary more than the lengths .