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प्रश्न
If for a distribution, `sum(x - 5)` = 3, `sum(x - 5)^2` = 43, and total number of observations is 18, find the mean and standard deviation
बेरीज
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उत्तर
`sum(x - 5)` = 3
⇒ `sumx - sum5` = 3
`sumx - 5sum1` = 3 ...[Note `sum5` = 5 × n]
`sumx - 5 xx 18` = 3
⇒ `sumx` = 3 + 90
`sumx` = 93
`bar(x)` = `(sumx)/"n"`
= `93/18` ......(1)
= 5.17
`sum(x - 5)^2` = 43
`sum(x^2 +25 - 10x)` = 43
⇒ `sumx^2 + sum25 - sum10x` = 43
`sumx^2 + 25 xx 18 - 10 xx 93` = 43
⇒ `sumx^2 + 450 - 930` = 43
`sumx^2` = 43 + 930 − 450
`sumx^2` = 523
Standard deviation (σ) = `sqrt((sumx^2)/"n" - ((sumx)/"n")^2`
= `sqrt(523/18 - (93/18)^2`
= `sqrt(29.06 - (5.17)^2`
= `sqrt(29.06 - 26.73)`
= `sqrt(2.23)`
= 1.53
(i) Arithmetic mean `(barx)` = 5.17
(ii) Standard deviation (σ) = 1.53
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