## The sum of deviations of a set of values x1 , x2 , x3 , –.. , xn , measured from 50 is -10 and the sum of deviations of the values fr

Question

The sum of deviations of a set of values x1 , x2 , x3 , ………. , xn , measured from 50 is -10 and the sum of deviations of the values from 46 is 10. find the value of n and the mean

0

1. n = 5

mean = 48

Let’s create a series of equations to express “sum of deviations of a set of values x1 , x2 , x3 , ………. , xn , measured from 50 is -10”

(x1 – 50) + (x2 – 50) + … + (xn – 50) = -10

(x1 + x2 + … + xn) – 50n = -10

x1 + x2 + … + xn = 50n – 10

And do the same for 46.

(x1 – 46) + (x2 – 46) + … + (xn – 46) = 10

(x1 + x2 + … + xn) – 46n = 10

x1 + x2 + … + xn = 46n + 10

Both 50n – 10 and 46n + 10 is equal to the sum of x1, x2, …, xn. So set them equal to each other and solve for n:

50n – 10 = 46n + 10

50n = 46n + 20

4n = 20

n = 5

Now we can calculate the sum of x1, x2, …, xn:

50n – 10 = 50*5 – 10 = 250 – 10 = 240

And the mean
mean = 240/5 = 48