Question

GIVING 99 PTS!!!
PLZ HELP FINAL EXAM QUESTION!!!
What is the mean absolute deviation for 2, 9, 1, 7, 8, and 9?

Answer 1

step 1 :- find mean of given numbers
mean (x') = (sum of numbers)/total number
= (2 + 9 + 1 + 7 + 8 + 9)/6
= 36/6
= 6

step 2 :-

|∆x1 |= x1 - x' = 2 - 6 = 4
|∆x2| = x2 - x' = 9 - 6 = 3
|∆x3| = x3 - x' = 1 - 6 = 5
|∆x4|= x4 - x' = 7 -6 = 1
|∆x5| = x5 - x' = 8 - 6 = 2
|∆x6 |= x6 - x' = 9 - 6 = 3

step 3 :-
take mean of |∆x1|.
, | ∆x2| ......|∆x6| , results is Mean Absolute Deviation.

MAD =( |∆x1 |+ |∆x2|+.....|∆x6|)/6
= ( 4 + 3 + 5 + 1 + 2 + 3)/6
= 3
hence, mean absolute deviation of given numbers = 3

Answer 2

Total Inputs(N) =(2, 9, 1, 7, 8, 9)

Total Inputs(N)=6

Mean(xm)= (x1+x2+x3...xN)/N

Mean(xm)= 36/6

Means(xm)= 6

-------------------------------------------

MAD=

1/(N)*(|x1-xm|+|x2-xm|+..+|xN-xm|)

=1/6(|2-6|+|9-6|+|1-6|+|7-6|+|8-6|+|9-6|)

=1/6(4+3+5+1+2+3))

=3