Q. 22

# 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.

Given, = -90

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

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

= ∑x – 15n = - 90 (I)

And, = 54

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

= (x1 + x2 + …. + xn) + (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 = = = Rate this question :

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