Q. The, coefficient of SD and coefficient of variance from the given data is [table] Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 ; Frequency | 2 | 10 | 8 | 4 | 6 [/table]

A) 50, 48.1
B) 51.9, 48.1
C) 0.481, 48.1
D) 0.481, 51.8
Explanation

(c)
[table] Class | Mid value x | $\mathbf{f}$ | $\mathbf{fx}$ | $\mathbf{d = x - m}$ | $\mathbf{fd}$ | $\mathbf{f}\mathbf{d}^{\mathbf{2}}$ ; 0-10 | 5 | 2 | 10 | -20.7 | -41.4 | 856.98 ; 10-20 | 15 | 10 | 150 | -10.7 | -107 | 1148.9 ; 20-30 | 25 | 8 | 200 | -0.7 | -5.6 | 3.92 ; 30-40 | 35 | 4 | 140 | 9.3 | 37.2 | 345.96 ; 40-50 | 45 | 6 | 270 | 19.3 | 115.8 | 2234.94 ; | | $\Sigma f = 30$ | $\Sigma fx = 770$ | | $\Sigma fd = - 1$ | $\Sigma fd^{2} = 4586.70$ [/table]
M=$\frac{770}{30} = 25.7$
SD ($\sigma$)=$\sqrt{\frac{\Sigma fd^{2}}{\Sigma f} - \left( \frac{\Sigma fd}{\Sigma f} \right)^{2}}$
$= \sqrt{\frac{4586.7}{30} - \left( \frac{- 1}{30} \right)^{2}}$
$\sqrt{15289 - 0.005 = 12365}$
$\therefore$Coefficient of SD=$\frac{\sigma}{x} = \frac{12.365}{25.7} = 0.481$
$and\ Coefficient\ of\ variance = coeff.of\ SD \times 100$
$= 0.481 \times 100 = 48.1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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