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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

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प्रश्न

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.

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उत्तर

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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अध्याय 15: Statistics - Exercise [पृष्ठ २८१]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 15 Statistics
Exercise | Q 22 | पृष्ठ २८१

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