Find the mean deviation about the mean for the data. \[38\], \[70\], \[48\], \[40\], \[42\], \[55\], \[63\], \[46\], \[54\], \[44\] - Noon Academy

Find the mean deviation about the mean for the data.

$38$

,

$70$

,

$48$

,

$40$

,

$42$

,

$55$

,

$63$

,

$46$

,

$54$

,

$44$

Class 11 | Exercise 15.1 | Maths | NCERT | Statistics

Find the mean deviation about the mean for the data.

$38$

,

$70$

,

$48$

,

$40$

,

$42$

,

$55$

,

$63$

,

$46$

,

$54$

,

$44$

Solution:-

To find mean deviation, first we have to find mean

$(\overline{x})$

$\overline{x}=\frac{1}{10}\sum\limits_{i=1}^{10}{{{x}_{i}}}=\frac{500}{10}=50$

Determine the respective values of the deviations from mean,

i.e.,

${{x}_{i}}-\overline{x}$

are,

$50-38=-12$

,

$50-70=-20$

,

$50-48=2$

,

$50-40=10$

,

$50-42=8$

,

$50-55=-5$

,

$50-63=-13$

,

$50-46=4$

,

$50-54=-4$

,

$50-44=6$

The deviations are

$-12,20,-2,-10,-8,5,13,-4,4,-6$

Therefore, the absolute values of the deviations,

$12,20,2,10,8,5,13,4,4,6$

$\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-\overline{x} \right|}=84$

We know that Mean deviation = sum of deviations/ number of observations

=

$84/10$

=

$8.4$

Hence, the mean deviation for the given data is

$8.4$

i

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