CBSE (Arts) Class 11CBSE
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# Variance and Standard Deviation - Shortcut Method to Find Variance and Standard Deviation

#### notes

By using step-deviation method, it is possible to simplify the procedure.
Let the assumed mean be ‘A’ and the scale be reduced to 1/h times (h being the width of class-intervals). Let the step-deviations or the new
values be y_i.
i.e. y_i = (x_i - A) /h  or x_i = A +hy_i                    ...(1)
we know that $\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} f_ix_i}{N}$

Replacing xi from (1) in (2), we get

$\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} f_i(A + hy_i)}{N}$
=$\frac {1}{N} (\displaystyle\sum_{i=1}^{n} f_i A + = \displaystyle\sum_{i=1}^{n} hf_iy_i )$ = =$\frac {1}{N} (A \displaystyle\sum_{i=1}^{n} f_i + h = \displaystyle\sum_{i=1}^{n} f_i y_i)$

=A .$\frac{N}{N} + h \frac {\displaystyle\sum_{i=1}^{n} f_iy_i}{N}$

(because $\displaystyle\sum_{i=1}^{n} f_i$ = N )

Thus bar x = A + h bar y         ...(3)

Now Variance of the variable x, $\sigma_x^2 =\frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i (x_i - \bar x)^2$

=$\frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i (A +hy_i - A - h \bar y)^2$  (Using (1) and (2))

=$\frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i h^2 (y_i - \bar y)^2$

= $\frac {h^2}{N} \displaystyle\sum_{i=1}^{n} f_i (y_i -\bar y)^2 = h^2 ×$ variance of the variable yi

i .e.sigma_x^2 = h^2 sigma_x^2

or σ  _x = hσ _y                        ...(4)
From (3) and  (4), we have

σx  = ${\frac{h}{N}}\sqrt{N\displaystyle\sum_{i=1}^{n} f_i y_i ^2 - (\displaystyle\sum_{i=1}^{n} f_i y_i) ^2 }$  ...(5)

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Shortcut method for Standard Deviation [00:09:49]
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