Mean Deviation - Free MCQ Practice Test with solutions, JEE Maths

Maximum and minimum value of the data 23, 45, 34, 21, 89, 45, 47, 91 are 21 and 91.
Range = 91 – 21 = 70
 

let, X =5,6,7,8,9,13,12,15.
(5+6+7+8+9+13+12+15)÷9 = 9.
and hence a = 9.
the mean deviation about the mean is summation of |X-a|÷ the total number
i.e , |X-a| = 4,3,2,1,3,0,4,3,6 and the total no. is 9.
hence summation of |X-a| = 26,
the mean deviation is 26 ÷ 9 = 2.89 ans

Formula

Fact

Here N= 5 , sigma x = 75
so mean = 15
now taking deviation from mean,( By ignoring signs)
we get sigma deviation from mean = 6
Now applying the formula of mean deviation
M.D.= SIGMA deviation from mean/ n
so M.D.= 6/5
= 1.2

Arrange data in ascending order,
3,4,6,7,8,8,9,10,12,13
No. of observations = 10
Median = n/2  => 10/2 = 5h observation.
5th observation is 8
Now we calculate mean deviation about median, i.e;
=> ∑∣xi−M∣/10
= {|3-8| +|4-8| +|6-8| +|7-8| +|8-8| +|8-8| +|9-8| +|10-8| +|12-8| +|13-8| }/10
= { 5 + 4 + 2 + 1 + 0 + 0 + 1 + 2 + 4 + 5}/10
= 24/10  => 2.4

Mean deviation of a data set is the average of the absolute deviations from a central point (Average value).