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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

The diameter of circles (in mm) drawn in the design are given below. Diameters 33 − 36 37 − 40 41 − 44 45 − 48 49 − 52 Number of circles 15 17 21 22 25 Calculate the standard deviation. - Mathematics

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Question

The diameter of circles (in mm) drawn in the design are given below.

Diameters 33 − 36 37 − 40 41 − 44 45 − 48 49 − 52
Number of circles 15 17 21 22 25

Calculate the standard deviation.

Sum

Solution

Assumed mean = 42.5

Diameters Number
of Circles
(fi)
Mid value
xi
d = xi − A
= xi − 42.5
fidi fidi2
33 − 36 15 34.5 − 8 − 120 960
37 − 40 17 38.5 − 4 − 68 272
41 − 44 21 42.5 0 0 0
45 − 48 22 46.5 4 88 352
49 − 52 25 50.5 8 200 1600
  `sumf_"i"` = 100     `sumf_"i""d"_"i"` = 100 `sumf_"i""d"_"i"^2` = 3184

Here `sumf_"i"` = 100, `sumf_"i""d"_"i"` = 100, `sumf_"i""d"_"i"^2` = 3184

Standard deviation (σ) = `sqrt((sumf_"i""d"_"i"^2)/(sumf_"i") - ((sumf_"i""d"_"i")/(sumf_"i"))^2`

= `sqrt(3184/100 - (100/100)^2`

= `sqrt(31.84 - 1)`

σ = `sqrt(30.84)`

Standard deviation (σ) = 5.55

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Measures of Dispersion
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Chapter 8: Statistics and Probability - Unit Exercise – 8 [Page 332]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 8 Statistics and Probability
Unit Exercise – 8 | Q 2 | Page 332

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