Question

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Further, another set of 15 observations x₁, x₂, ..., x₁₅, also in seconds, is now available and we have `sum_(i = 1)^15 x_i` = 279 and `sum_(i  = 1)^15 x^2` = 5524. Calculate the standard derivation based on all 40 observations.

Answer 1

Given, n₁ = 25

`barx_1` = 18.2

`sigma_1` = 3.25

n₂ = 15

`sum_(i = 1)^15 x_i` = 279

And `sum_(i = 1)^15 x_i^2` = 5524

For first set `sumx_i = 25 xx 18.2` = 455

∴ `sigma_1^2 = (sumx_i^2)/25 - (18.2)^2`

⇒ `(3.25)^2 = (sumx_i^2)/25 - (18.2)^2`

⇒ 10.5625 + 331.24 = `(sumx_i^2)/25`

⇒ `sumx_i^2 = 25 xx (10.5625 + 331.24)`

= `25 xx 341.8025`

= 8545.0625

For combined S.D. of the 40 observations, n = 40.

Now `sum_(i = 1)^40 x_i^2` = 5524 + 8545.0625 = 14069.0625

And `sum_(i = 1)^40 x_i` = 455 + 279 = 734

∴ S.D. = `sqrt((14069.0625)/40 - (734/40)^2`

= `sqrt(351.1726 - (18.35)^2`

= `sqrt(351.726 - 336.7225)`

= `sqrt(15.0035)`

= 3.87