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: Which is more varying, the length or weight?

Answer 1

The given data is,

∑ i=1 50 x i =212 , ∑ i=1 50 ( x i ) 2 =902.8 , ∑ i=1 50 y i =261 and ∑ i=1 50 ( y i ) 2 =1457.6 .

It is clear that the value of N is 50 as the values are terminating up to 50.

The formula to calculate the mean is,

x ¯ = ∑ i=1 n x i N (1)

Substitute ∑ i=1 50 x i =212 and N=50 in equation (1).

x ¯ = 212 50 =4.24

The formula to calculate the variance is,

σ 2 = 1 N ∑ i=1 n ( x i − x ¯ ) 2 (2)

Substitute 4.24 for x ¯ and 50 for N in equation (2).

σ 2 = 1 50 ∑ i=1 n ( x i −4.24 ) 2

Simplify the above equation by using square formula and applying limits to each term.

σ 2 = 1 50 [ ∑ i=1 n ( x i ) 2 − ∑ i=1 n 8.48 x i + ∑ i=1 n 17.97 ]

Since the terms are up to 50 thus the equation becomes,

σ 2 = 1 50 [ ∑ i=1 n ( x i ) 2 −8.48 ∑ i=1 n x i +17.97( 50 ) ]

Substitute ∑ i=1 50 ( x i ) 2 =902.8 and ∑ i=1 50 x i =212 in the above equation.

σ 2 = 1 50 [ 902.8−8.48( 212 )+17.97( 50 ) ] = 1 50 ( 1801.3−1797.76 ) = 1 50 ( 3.54 ) =0.07

The formula to calculate the standard deviation is,

S.D.= σ 2 (3)

Substitute 0.07 for σ 2 in equation (3).

S.D.= 0.07 =0.26

The formula to calculate the coefficient of variation is,

C.V.= σ x ¯ ×100 (4)

Substitute 0.26 for σ and 4.24 for x ¯ in equation (4).

C.V.= 0.26 4.24 ×100 =6.13

Thus, the coefficient of variation for length is 6.13.

Similarly, calculate the values for weight.

Substitute ∑ i=1 50 y i =261 and N=50 in equation (1).

y ¯ = 261 50 =5.22

Substitute 5.22 for y ¯ and 50 for N in equation (2).

σ 2 = 1 50 ∑ i=1 n ( y i −5.22 ) 2

Simplify the above equation using the square formula and applying limits to each term.

σ 2 = 1 50 [ ∑ i=1 n ( y i ) 2 − ∑ i=1 n 10.44 y i + ∑ i=1 n 27.24 ]

Since the terms are up to 50, the equation becomes,

σ 2 = 1 50 [ ∑ i=1 n ( y i ) 2 −10.44 ∑ i=1 n y i +27.24( 50 ) ]

Substitute ∑ i=1 50 ( y i ) 2 =1457.6 and ∑ i=1 50 y i =261 in the above equation.

σ 2 = 1 50 [ 1457.6−10.44( 261 )+1362 ] = 1 50 ( 2819.6−2724.84 ) = 1 50 ( 94.76 ) =1.89

Substitute 1.89 for σ 2 in equation (3).

S.D.= 1.89 =1.37

Substitute 1.37 for σ and 5.22 for y ¯ in equation (4).

C.V.= 1.37 5.22 ×100 =26.24

Thus, the coefficient of variation for weight is 26.24.

Since the coefficient of variation for weights is more than the lengths, therefore weights vary more than the length.

Thus, weights vary more than the lengths.