The diameters of the 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]

Solution:

Let us draw the frequency table based on the given problem

Class interval	Frequency fi	Mid-point (xi)	yi = (xi - 42.5)/4	yi²	fiyi	fiyi²
33 - 36	15	34.5	- 2	4	- 30	60
37 - 40	17	38.5	- 1	1	- 17	17
41 - 44	21	42.5	0	0	0	0
45 - 48	22	46.5	1	1	22	22
49 - 52	25	50.5	2	4	50	100
	100				25	199

Here,

N = 100, h = 4

Let the assumed mean,

A be 42.5

Mean,

x = A + [∑⁵_{i = 1} f_iy_i]/N × h

= 42.5 + 25/100 × 4

= 42.5 + 1

= 43.5

Variance,

(σ ²) = h²/N² [N ∑⁵_{i = 1}(f_iy_i)² - (∑⁵_{i = 1}f_iy_i)²]

= (16)/(10000) [100 × 199 - (25)²]

= (16)/(10000) [19900 - 625]

= (16)/(10000) × 19275

= 30.84

Standard deviation,

(σ) = √ 30.84

= 5.55

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NCERT Solutions Class 11 Maths Chapter 15 Exercise 15.2 Question 10

The diameters of the 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]

Summary:

The mean, variance, and standard deviation for the given data are 43.5, 30.84, and 5.55 respectively