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Question
Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Solution
Given that `n = 100, barx = 40, sigma = 10`
∴ `barx = (sumx_i)/N`
⇒ 40 = `(sumx_i)/100`
⇒ `sumx_i` = 400
Corrected `sumx_i` = 4000 – 30 – 70 + 3 + 27 = 3930
And Corrected mean = `3930/100` = 39.3
Now `sigma^2 = (sumx_i^2)/n - (40)^2`
⇒ 100 = `(sumx_i^2)/100 - 1600`
⇒ `sumx_i^2 = 1700 xx 100`
⇒ `sumx_i^2` = 170000
∴ Corrected `sumx_i^2 = 17000 - (30)^2 - (70)^2 + (3)^2 + (27)^2`
= 170000 – 900 – 4900 + 9 + 729
= 164938
∴ Correct S.D. = `sqrt(164938/100 - (39.3)^2`
= `sqrt(1649.38 - 1544.49)`
= `sqrt(104.89)`
= 10.24
Hence, the required S.D. = 10.24
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