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The diameters of circles (in mm) drawn in a design are given below: Diameters 33 - 36 37 - 40 41 - 44 45 - 48 49 - 52 No. of circles 15 17 21 22 25 - Mathematics

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Question

The diameters of circles (in mm) drawn in a design are given below:

Diameters 33 - 36 37 - 40 41 - 44 45 - 48 49 - 52
No. of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.

[Hint: First make the data continuous by making the classes as 32.5 - 36.5, 36.5 - 40.5, 40.5 - 44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.]

Sum
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Solution

The intervals to convert the given discrete data into continuous frequency distribution are as follows.

32.5 - 36.5, 36.5 - 40.5, 40.50 - 44.5, 44.5 - 48.5, 48.5 - 52.5

Let A = 42.5, h = 4,

∴ yi = `(x_i - 42.5)/4`

Class Interval Mid-point Frequency yi fiyi yi2 fiyi2
32.5 - 36.5 34.5 15 −2 −30 4 60
36.5 - 40.5 38.5 17 −1 −17 1 17
40.50 - 44.5 42.5 21 0 0 0 0
44.5 - 48.5 42.5 22 1 22 1 22
48.5 - 52.5 50.5 25 2 50 4 100
Sum - 100 - 25 - 199

Mean, `overlinex = A +((sumf_iy_i)/N) xx h`

= `42.5 + 25/100 xx 4`

= 42.5 + 1

= 43.5

Variance σ2 = `h^2/N^2[Nsumf_iy_i^2 - (sumf_iy_i)^2]`

= `16/(100)^2 [100 xx 199 - (25)^2]`

= `(16 xx 25)/(100 xx 100) [4 xx 199 - 25]`

= `1/25 [796 - 25]`

= `771/25`

= 30.84

∴ Standard deviation σ = `sqrt30.48` = 5.56

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Chapter 13: Statistics - EXERCISE 13.2 [Page 282]

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NCERT Mathematics [English] Class 11
Chapter 13 Statistics
EXERCISE 13.2 | Q 10. | Page 282

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