The sums of the deviations of a set of n values x 1 , x 2 , x 3 ,…., x n measured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.

Question

The sums of the deviations of a set of n values x 1 , x 2 , x 3 ,&hellip;., x n measured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.

Philoid · Accepted Answer

Given, = -90

= (x₁ – 15) + (x₂ – 15) + …. + (x_n – 15) = - 90

= (x₁ + x₂ + … + x_n) – (15 + 15 + … + 15) = -90

= ∑x – 15n = - 90 (I)

And, = 54

= (x₁ + 3) + (x₂ + 3) + …. + (x_n + 3) = 54

= (x₁ + x₂ + …. + x_n) + (3 + 3 + …. + 3) = 54

= ∑x + 3n = 54 (II)

Subtracting (I) from (II), we get

∑x + 3n - ∑x + 15n = 54 + 90

18n = 144

n =

= 8

Put value of n in (I), we get

∑x – 15 * 8 = - 90

∑x – 120 = - 90

∑x = 30

Therefore, Mean =

=

The sums of the deviations of a set of n values x1, x2, x3,…., xnmeasured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.

The sums of the deviations of a set of n values x₁, x₂, x₃,…., x_nmeasured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.