Given set m: 0, 1, 3, 4, 7 and set n: 1, 2, 3, 4, 5, 9. calculate the mean and mean absolute deviation for both sets. then, answer the following questions. how many data points in set m are closer than one mean absolute deviation from the mean of set m? how many data points in set n are further than two mean absolute deviations from the mean of set n? find the difference between the two means. how does this difference relate to the mean absolute deviation of each set? for example, is it the same as the mean absolute deviation, twice as much, half as much, etc?

solution:

set m : 0, 1, 3, 4, 7.

set n : 1, 2, 3, 4, 5, 9

mean of set m =

mean of set n =  

absolute difference between data values of set m and mean of set m

  = 3, 2, 0, 1, 4

absolute difference between data values of set m and mean of set m

  = 3,2,1,0,1,5

mean absolute deviation of set m =

number of   data points in set m which are closer than one mean absolute deviation from the mean of set m = 2 which are 1,3 because 2-1=1 and 2 +1=3, equidistant from 2.

mean absolute deviation of set n =

number of   data points in set n which are closer than one mean absolute deviation from the mean of set n = 3, which are 1,2,3 because 2-1=1, 2+1=3, and 2-0 or 2 +0 is equal to 2.

(b) difference between the two means=

  [mean of set m - mean of set n] or [mean of set n - mean of set m ]

that is 3-4 or 4-3

  = -1 or 1

absolute   difference between the two means= 1

(c) : absolute   difference between the two means= 2=1,

=     [mean absolute deviation of set m or set n]

→→mean absolute deviation of set m or set n=2 × absolute   difference between the two means

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

1/4 because there is 4 grades.