Answer :
Answer:
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
Answer:
See below
Step-by-step 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,
[tex] \rm\displaystyle \: MAD = \frac{1}{n} \sum_{i = 1} ^{n} |x_{i} - m(x)| [/tex]
where m(X) represents the mean of the given database and [tex]|x_{i}-m(X)|[/tex] tells us to subtract every data with the got mean and n represents the number of data
let figure out m(X)
[tex] \rm\displaystyle \: m(x) = \frac{3 + 4 + 5(8) + 6 + 7 }{12} [/tex]
simplify multiplication:
[tex] \rm\displaystyle \: m(x) = \frac{3 + 4 + 40 + 6 + 7 }{12} [/tex]
simplify addition:
[tex] \rm\displaystyle \: m(x) = \frac{60}{12} [/tex]
simplify substraction:
[tex] \rm\displaystyle \: m(x) = 5[/tex]
let's figure out [tex]\sum[/tex] and [tex]|x_i-m(X)|[/tex]
[tex] \displaystyle\begin{array} {|c |c| } \hline |x_{i} - m(x) | & \displaystyle\sum_{i = 1} ^{n} |x_{i} - m(x) | \\ \hline |3 - 5 | & 2 \\ \hline | 4 - 5 | &1 \\ \hline |5(8) - 5(8)| &0 \\ \hline |6 - 5|&1 \\ \hline |7 - 5| &2 \\ \hline \\ \text{total} & 5 \\ \hline\end{array}[/tex]
last part let's figure out MAD
[tex] \displaystyle \: MAD = \frac{5}{12} [/tex]
simplify division:
[tex] \displaystyle \: MAD = 0.42[/tex]
Class-2:
likewise class-2
given database 7,8,9,9,9,9,9,9,9,9,10,11
figure out m(X)
[tex] \rm\displaystyle \: m(x )= \frac{7 + 8 + 9(8) + 10 + 11}{12} [/tex]
simplify multiplication:
[tex] \rm\displaystyle \: m(x )= \frac{7 + 8 + 72 + 10 + 11}{12} [/tex]
simplify addition:
[tex] \rm\displaystyle \: m(x )= \frac{108}{12} [/tex]
simplify division:
[tex] \rm\displaystyle \: m(x )= 9[/tex]
likewise
figure out [tex]\sum[/tex] and [tex]|x_i-m(X)|[/tex]
[tex] \displaystyle\begin{array} {|c |c| } \hline |x_{i} - m(x) | & \displaystyle\sum_{i = 1} ^{n} |x_{i} - m(x) | \\ \hline |7 - 9 | & 2 \\ \hline | 8- 9 | &1 \\ \hline |9(8) - 9(8)| &0 \\ \hline |10 - 9|&1 \\ \hline |11- 9| &2 \\ \hline \\ \text{total} & 5 \\ \hline\end{array}[/tex]
figure out MAD
[tex] \displaystyle \: MAD = \frac{5}{12} [/tex]
simplify division:
[tex] \displaystyle \: MAD = 0.42[/tex]