Question

For the data set $\{10,12,40,35,14,24,13,21,42,30\}$, find the mean absolute deviation to the nearest hundredth.

• A. $9.12$
• B. $11.12$
• C. $12.12$
• D. $10.12$

Answer 1

The correct option is D $10.12$

											SUM
$X$	10	12	40	35	14	24	13	21	42	30	241
$|X-\overline{X}|$	+14.1	+12.1	+15.9	10.9	+10.1	+0.1	+11.1	+3.1	17.9	5.9	101.20
$(X-\overline{X}{)}^2$	198.81	146.41	252.81	118.81	102.01	0.01	123.21	9.61	320.41	34.81	1306.9

$\overline{X}=\frac{\sum X}{N}=\frac{241}{10}=24.1$

${\sigma }_x=\sqrt{\frac{\sum (X-\overline{X}{)}^2}{N}}=\sqrt{130.69}=11.43$

Variance $={\sigma }_x^2=130.69$

New absolute Deviation $=\frac{\sum |X_i-\overline{X}|}{N}=\frac{101.20}{10}=10.12$