The sums of the deviations of a set of `n` values `x_1, x_2,ddot,x_n` measured from `15` and `-3a r e-90` and `54` respectively. Find the value of `n` and mean.

Explanation

Solution
The sum of the deviations of set of `n` values:
`x_1,x_2,x_3,.........x_n`
Measured from `15 `and `−3` are `−90` and 54 respectively
Sum of the deviations of set of `n` value `x_1,x_2,x_3,.........x_n`
​Measured from `15` are `−90`
i.e `sum_(i=0)^N (xi−15) =−90............(1)`
And,
Sum of the deviations of set of n value `x1,x2,x3,.........xn `
Measured from`−3` are` 54`
`sum_(i=1)^N (xi −(−3))=54`
`sum_(i=1)^N(xi+3)=54`............`(2)`
`(1)` and `(2)`
`sum_(i=1)^N​(xi−15)=−90`
`=> sum_(i=1)^N xi−15 sum_(i=1)^N 1=−90`
` n bar x −15n=−90.............(3)`
`sum_(i=1)^N (xi +3)=54`
`=> sum_(i=1)^N xi+3 sum_(i=1)^N 1=54`
`n bar x +3n=54.............(4)`
We know
`bar x sum_(i=1)^N xi/n`
`n bar x sum_(i=1)^N xi` and `sum_(i=1)^N = n`
`(4)-(3)`
`=>n bar x + 3n−n bar x+15n=54+90=144`
`18n=144`
`=> n = 144/18 = 8`
`n=8`
`=>` eq `(4)`
`=> n bar x + 3n = 54`
`=>8(bar x + 3)= 54`
`= bar x + 3=54/8`
`= 54/8 - 3`
`=(54−24)/8`
`= 30/8=15/4`
`bar x=15/ 4`
Hence the values of n and mean are `8` and `15/4`