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Measures of Dispersion — Economics STD 11 Commerce — Question

CBSE BoardEnglish MediumSTD 11 CommerceEconomicsMeasures of Dispersion6 Marks
Question
Find the standard deviation by the step deviation method.
Class lnterval: $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency (f): $16$ $12$ $10$ $15$ $12$ $8$
✓

Answer

Calculation of Standard Deviation
Class lnterval frequency$(f)$ Mid value $(m)$ $d(m - A)A = 25$ $\text{d}'\Big(\frac{\text{d}}{\text{c}}\Big)\text{c}=10$ $d'^2$ $fd'$ $fd'^2$
$0-10$ $16$ $5$ $-20$ $-2$ $4$ $-32$ $64$
$10-20$ $12$ $15$ $-10$ $-1$ $1$ $-12$ $12$
$20-30$ $10$ $25$ $0$ $0$ $0$ $0$ $0$
$30-40$ $15$ $35$ $10$ $1$ $1$ $15$ $15$
$40-50$ $12$ $45$ $20$ $2$ $4$ $24$ $48$
$50-60$ $8$ $55$ $30$ $3$ $9$ $24$ $72$
  $\Sigma\text{f}=73$         $\Sigma\text{fd}'=19$ $\Sigma\text{fd}'^2=211$
$\sigma\sqrt{\frac{\Sigma\text{fd}'^2}{\Sigma\text{f}}-\Big(\frac{\Sigma\text{fd}'}{\Sigma\text{f}}\Big)^2}\times\text{c}$
$=\sqrt{\frac{211}{73}-\Big(\frac{19}{73}\Big)^2}\times10$
$=\sqrt{2.89-0.067}\times10$
$=\sqrt{2.82}\times10=1.68\times10$
$\therefore\sigma=16.8$

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