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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएसएसएलसी (अंग्रेजी माध्यम) कक्षा १०

If for a distribution, ∑(x-5) = 3, ∑(x-5)2 = 43, and total number of observations is 18, find the mean and standard deviation - Mathematics

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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

योग

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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Measures of Dispersion
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Statistics and Probability - Unit Exercise – 8 [पृष्ठ ३३२]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
अध्याय 8 Statistics and Probability
Unit Exercise – 8 | Q 5 | पृष्ठ ३३२

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