Question

The line plots below show the numbers of hours for the 12 students in each class.

The distance between the means of Class A and Class B is 4 hours.
Find the mean absolute deviation (MAD) for each class.

Answer 1

See below

Explanation:

class-1:

we are given some data 3,4,5,5,5,5,5,5,5,5,6,7

we want to figure out the MAD (Mean absolute deviation) of the given database

remember that,

`\rm\displaystyle \: MAD = (1)/(n) \sum_(i = 1) ^(n) |x_(i) - m(x)|`

where m(X) represents the mean of the given database and
`|x_(i)-m(X)|` tells us to subtract every data with the got mean and n represents the number of data

let figure out m(X)

`\rm\displaystyle \: m(x) = (3 + 4 + 5(8) + 6 + 7 )/(12)`

simplify multiplication:

`\rm\displaystyle \: m(x) = (3 + 4 + 40 + 6 + 7 )/(12)`

simplify addition:

`\rm\displaystyle \: m(x) = (60)/(12)`

simplify substraction:

`\rm\displaystyle \: m(x) = 5`

let's figure out
`\sum` and
`|x_i-m(X)|`

last part let's figure out MAD

`\displaystyle \: MAD = (5)/(12)`

simplify division:

`\displaystyle \: MAD = 0.42`

Class-2:

likewise class-2

given database 7,8,9,9,9,9,9,9,9,9,10,11

figure out m(X)

`\rm\displaystyle \: m(x )= (7 + 8 + 9(8) + 10 + 11)/(12)`

simplify multiplication:

`\rm\displaystyle \: m(x )= (7 + 8 + 72 + 10 + 11)/(12)`

simplify addition:

`\rm\displaystyle \: m(x )= (108)/(12)`

simplify division:

`\rm\displaystyle \: m(x )= 9`

likewise

figure out
`\sum` and
`|x_i-m(X)|`

figure out MAD

`\displaystyle \: MAD = (5)/(12)`

simplify division:

`\displaystyle \: MAD = 0.42`

Answer 2

for class A

mean ={3+4+8(5)+6+7)/12}=5hours

mean absolute deviation (MAD) for A

=[|5-3|+|5-4|+8|5-5|+|5-6|+|5-7|]/12=6/12=1/2

for class B

mean =4 hours given

mean absolute deviation (MAD) for B

=[|4-7|+|4-8|+8|4-9|+|4-10|+|4-11|]/12=60/12=5