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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.
Solution
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
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