Question

If for a distribution, `sum(x - 5)` = 3, `sum(x - 5)^2` = 43, and total number of observations is 18, find the mean and standard deviation

Answer 1

`sum(x - 5)` = 3

⇒ `sumx - sum5` = 3

`sumx - 5sum1` = 3   ...[Note `sum5` = 5 × n]

`sumx - 5 xx 18` = 3

⇒ `sumx` = 3 + 90

`sumx` = 93

`bar(x)` = `(sumx)/"n"`

= `93/18` ......(1)

= 5.17  

`sum(x - 5)^2` = 43

`sum(x^2 +25 - 10x)` = 43

⇒ `sumx^2 + sum25 - sum10x` = 43

`sumx^2 + 25 xx 18 - 10 xx 93` = 43

⇒ `sumx^2 + 450 - 930` = 43

`sumx^2` = 43 + 930 − 450

`sumx^2` = 523

Standard deviation (σ) = `sqrt((sumx^2)/"n" - ((sumx)/"n")^2`

= `sqrt(523/18 - (93/18)^2`

= `sqrt(29.06 - (5.17)^2`

= `sqrt(29.06 - 26.73)`

= `sqrt(2.23)`

= 1.53

(i) Arithmetic mean `(barx)` = 5.17

(ii) Standard deviation (σ) = 1.53