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प्रश्न
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.]
उत्तर
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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