Q. 22

# 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.

_{n}

Answer :

Given, = -90

= (x_{1} – 15) + (x_{2} – 15) + …. + (x_{n} – 15) = - 90

= (x_{1} + x_{2} + … + x_{n}) – (15 + 15 + … + 15) = -90

= ∑x – 15n = - 90 (I)

And, = 54

= (x_{1} + 3) + (x_{2} + 3) + …. + (x_{n} + 3) = 54

= (x_{1} + x_{2} + …. + 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 =

=

=

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