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If for distribution ∑(x-5) = 3, ∑(x-5)2 = 43 and total number of items is 18. Find the mean and standard deviation.

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प्रश्न

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

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उत्तर

Given that `n = 18, sum(x - 5) = 3, sum(x - 5)^2` = 43

∴ Mean = `A + (sum(x - 5))/n`

= `5 + 3/18`

= `93/18`

= 5.166

= 5.17

And S.D. = `sqrt((sum(x - 5)^2)/N - [(sum(x - 5)^2)/N]^2`

= `sqrt(43/18 - (3/18)^2`

= `sqrt(2.39 - (0.166)^2`

= `sqrt(2.39 - 0.27)`

= 1.54

Hence, the required mean is 5.17 and S.D. = 1.54

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अध्याय 15: Statistics - Exercise [पृष्ठ २७९]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 15 Statistics
Exercise | Q 14 | पृष्ठ २७९

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