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Mathematics Question on Variance and Standard Deviation

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

DiametersNo. of children
33-3615
37-4017
41-4421
45-4822
49-5225
Answer
Class IntervalFrequency fif_imidpointximid-point\,x_iyi=xi42.54y_i=\frac{x_i-42.5}{4}fi2f_i^2fiyif_iy_ifiy12f_iy_1^2
32.5-36.51534.5-24-3060
36.5-40.51738.5-11-1717
40.5-44.52142.50000
44.5-48.52246.5112222
48.5-52.52550.52450100
10025199

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5

Mean, xˉ=Ai=15fixin×h=42.5+25100×4=43.5\bar{x}=A\frac{\sum_{i=1}^5f_ix_i}{n}×h=42.5+\frac{25}{100}×4=43.5

Variance (σ2) = h2N2[Ni=15fiyi2(i=15fiyi)2]\frac{h^2}{N^2}[N\sum_{i=1}^5f_iy_i^2-(\sum_{i=1}^5f_iy_i)^2]

=1610000[100×199(25)2]=\frac{16}{10000}[100×199-(25)^2]

=1610000[19900625]=\frac{16}{10000}[19900-625]

=161000×19275=\frac{16}{1000}×19275

=30.84=30.84

Standarddeviation.(σ)=5.55Standard\,\,deviation\\.(σ) =5.55